LinkedIn Insight UD-TRG-Physics-Sem2-Ans-1Paper-3 - Grad Plus

UD-TRG-Physics-Sem2-Ans-1Paper-3

B.E. Second Semester All Branches (C.B.S.)
Time: 2 hours
Maximum marks: 60

Notes :
1.Question No. 1 is compulsory.
2. Attempt any three questions from Q.2 to Q.6.

  1. Used suitable data wherever required.
  2. Figures to the right indicate full mark.
  3. Attempt any five of the following :- [15M]
    (a) Comment on colours in a soap film in sunlight.

If sunlight is incident on a soap film, the optical path difference will vary from one color to other due to change of wavelength with color. Therefore the film will appear to be colored and the color seen will be of that rays for which constructive interference occurs. If the sunlight is not parallel then the optical path difference will change due to change in angle of incidence. Hence, the film will show different colors when viewed from different direction.

(b) What is Rayleigh’s criterion of resolution ? Define resolving power of a grating.

When an object is seen through a optical instrument, diffraction occurs as the light passes through the aperture. Hence in actual practice the diffraction pattern of the object is seen when viewed through any optical instrument. Because of which there is a limit on the ability of that instrument to view the two closely spaced point objects separately.

According to Rayleigh’s criterion the closely spaced point objects are said to be just resolved, or can be seen separately, when the principle maxima of one objects fall upon the first minimum of second object and vice versa. If the distance decreases then they are not resolved and if distance increased then they are said to be well resolved.

Resolving power of grating can be defined as the reciprocal of angular or linear distance for which two spectral lines can be seen separately and is given by

R=λ/Δλ.

(c) Calculate V number for an optical fiber having numerical aperture 0.25 and core diameter 20 μm if it is operated at 1.55 μm.

Given,
NA=0.25, d= 20 x 10-6 m, λo= 1.55 x 10-6 m,

We have,

\begin{array}{l}V=\frac{\pi d}\lambda\left[NA\right]\\\;\;\;=\frac{3.14\times20\times10^{-6}}{1.55\times10^{-6}}\times0.25\\\;\;\;=10.129\end{array}

(d) Compare light from ordinary source with laser light.

1) Light from ordinary source is not directional, whereas laser light is highly directional.

2) Ordinary light is poly-chromatic whereas LASER is highly monochromatic and contains a very
narrow range of the order of a few Angstroms (<10 Å ). 3) The intensity of ordinary light is very small whereas LASER emits light in the form of a narrow beam with its energy concentrated in a small region of space .Therefore, the beam intensity would be tremendously large and stays constant with distance since it travels in the form of plane waves. 4) Ordinary light is highly divergent whereas Light from a LASER propagates in the form of plane waves and it has very small divergence. 5) ordinary light is incoherent light since they emit light waves of random wavelengths having random phases. While, the waves emitted by LASER source are in phase and having same frequency. Therefore, light emitted by LASER is highly coherent. (e) How phase difference between two signals is measured using CRO? When two sine waves oscillating in mutually perpendicular direction are of same frequency, the Lissajous pattern takes the form of an ellipse as shown in the figure. This ellipse is used to determine the phase difference between two signals, by using the formula \phi=\sin^{-1}\left(\frac{AA'}{BB'}\right) By measuring the length AA’ and BB’ on the oscillogram and substituting the values in the above formula phase difference can be determined. Images……. (f) What are the properties of matter waves? 1. Wavelength of Matter waves depends upon mass and velocity of particle \left(\lambda=\frac h{m\nu}\right), where as wavelength of EM waves depends upon energy \left(\lambda=\frac{hc}E\right). 2. Velocity of matter wave depend upon velocity of particle and greater than velocity of light, whereas velocity of EM wave is constant equal to velocity of light. 3. Matter waves are not real waves but are probability waves only, whereas EM waves are real waves with periodic variation of E and B vector in them. 4. Matter waves are produced by a motion of particle and are independent of charges, whereas EM waves are produced by motion of charged particle only. (g) A superconductor has a critical temperature 3.7°K at zero magnetic field. At O°K the critical magnetic field is 0.0306 Tesla. What is the critical magnetic field at temperature 2.0°K? We have, \begin{array}{l}H_c\left(T\right)=H_c\left(o\right)\left[1-\left(\frac T{T_c}\right)\right]^2\\\;\;\;\;\;\;\;\;\;\;=0.0306\left[1-\left(\frac2{3.7}\right)\right]^2\\H_c\left(T\right)=0.00646\;K\\\end{array} 2. (a) Show that the diameter of Newton’s nth dark ring is proportional to square root of ring number. In Newton’s rings experiment the diameter of 5th dark ring was 0.336 cm and that of 15th dark ring was 0.590 cm. Calculate the radius of curvature of plano-convex lens if wavelength of light used is 5890 Å. [8M] Let R be the radius of curvature of the lens. Let a dark ring be located at the point B. The thickness of the air film at B is t and AE = r, the radius of the ring at thickness t. By using the Pythagoras theorem to triangle AEO we get, AO2 =AE2+EO2 R2 = r2+ (R-t)2 = r2 + R2 – 2Rt + t2 r2 = 2Rt or t = r2/2R  ,As t is very small, t2 can be neglected Using relation 2t = nλ, for dark rings. r2 = nλR r=\sqrt{n\lambda R} Dark ring. The Diameter of dark ring is given by D_n=\sqrt{4n\lambda R} Since, λ and  R are constants D_n\alpha\sqrt n i.e. the diameter of the Newton’s nth dark ring is directly proportional to square root of the ring number. we have,

\begin{array}{l}R=\frac{D_n^2}{4n\lambda}\\\;\;\;=\frac{\left(0.336\times10^{-2}\right)^2}{4\times5\times5.89\times10^{-7}}\\\;\;\;=0.000958\times10^3\\\;\;\;=95.8\times10^{-2}m\\\;\;\;=95.8\;cm\end{array}<a href="b">/latex</a> Derive an expression for numerical aperture of step index optical fiber. What are the advantages of using an optical fiber? [7M] The main function of the optical fiber is to accept and transmit as much as light from the source as possible. The light gathering ability of the fiber depends on two factors, namely core size and numerical aperture. The numerical aperture determined by the acceptance angle and fractional refractive index change. Acceptance angle and acceptance cone image…… Let us consider an optical fiber into which light is launched. Let refractive index of core be µ1 and refractive index of cladding be µ2.The refractive index of cladding is less than refractive index of core. Let µ0 be the refractive index of the medium from which light is launched into the fiber. Let a light ray enter the fiber at an angle θi to the axis of the fiber. The rays refracts at an angle θr and strike the core cladding interface at an angle [latex] \phi

. If \phi is greater than the critical angle \phic, the rays undergoes total internal reflection at the interface, since µ1>µ2 as long as the angle \phi is greater than critical angle \phic, the light will stay within the fiber as shown figure.

Let us now compute the incident angle θi for which  \phi\;\geq\;\phi_c such that light rebounds within the
fiber. Therefore From Snell’s law,

\begin{array}{l}\frac{\sin\theta_i}{\sin\theta_r}=\frac{\mu_1}{\mu_2}\\\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_2}\sin\theta_r-----\left(i\right)\end{array}

If θi is increased beyond a limit, \phi will drop below the critical value \phic and the ray escape from the sides walls of the fiber.

\sin\theta_r=\sin\left(90-\phi\right)=\cos\phi\;-----\left(ii\right)

Using equation 1 and 2

\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi

when \phi=\phi_c

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi_c-----\left(iii\right)

But by the condition of total internal reflection

\sin\phi_c=\frac{\displaystyle\mu_2}{\displaystyle\mu_1}

Now, Squaring both side

\begin{array}{l}\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\;;\;1-\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\\\cos^2\phi_c=\frac{\mu_1^2-\mu_2^2}{\mu_1^2}\;;\;\cos^2\phi_c=\frac{\displaystyle\mu_1^2-\mu_2^2}{\displaystyle\mu_1^2}\\\cos\phi_c=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_1}-----\left(iv\right)\end{array}

Put equation (iv) in equation (3), we get,

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_0}

Incident ray is launched from air medium for which, \mu_0=1 and substituting \theta_{i\left(max\right)}=\theta_0

we get,

\begin{array}{l}\sin\theta_0=\sqrt{\mu_1^2-\mu_2^2}\\\theta_0=\sin^{-1}=\sqrt{\mu_1^2-\mu_2^2}\end{array}
The angle θ0 is called the acceptance angle of the fiber.

Acceptance angle: It is defined as the maximum angle that a light ray can have relative to the axis of the fiber and propagate down the fiber.

Acceptance Cone: The light ray contained within the cone having a full angle 2θ0 are accepted and transmitted along the fiber. Therefore the cone is called the acceptance cone.

Fractional refractive index change
The fractional difference Δ between the refractive indices of the core and the cladding is known as Fractional refractive index change.

\triangle=\frac{\displaystyle\mu_1-\mu_2}{\mu_1}

Numerical aperture: The numerical aperture (NA) is defined as the acceptance angle.

Thus

\begin{array}{l}N.A.=\sin\theta_0\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}{2\mu_1}}2\mu_1\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1-\mu_2\right)}{\mu_1}}2\mu_1^2\;\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1+\mu_2}2=\mu_1\right)\\N.A.=\mu_1\sqrt{2\triangle}---\left(i\right)\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1-\mu_2}{\mu_1}=\triangle\right)\end{array}

Numerical aperture determines the light gathering ability of the fiber. It is measure of the amount of the light that can be accepted by fiber. It is seen from equation (i) that NA. dependent only on the refractive indices of the core and cladding materials. Its value range from 0.13 to 0.50. A large NA implies that a fiber will accept large amount of light from the source.

  1. (a) Explain construction and working of He-Ne laser. What are its merits ? [8M]

Construction: It consists of a glass discharge tube of about 30 cm long and 1 cm diameter. It is filled with mixture of He and Ne gas in the ratio of 10:1. Two electrodes are provided to produce discharge in the gas and connected to high voltage power supply. The tube is hermetically sealed by inclined windows arranged at its two ends. On the axis of the tube two reflectors are fixed which constitute optical resonator cavity. The distance between the mirrors is adjusted such that it supports standing wave pattern.

Image…..

Working:
The energy level diagram is shown in figure. It employs four level pumping scheme. When the power supply is put on; due to electric field the gas is ionized producing electrons and ions of gas. The electrons and ions are accelerated towards anode and cathode. The electrons transfer their energy to He ions due to collision and are excited to higher state which lie at 19.81 eV and 20.16 eV above the ground state. He- atoms are lighter and hence readily excited. These levels E2 and E3 are metastable states and hence the excited He atoms do not return to ground states through spontaneous emission process. However, the He atoms return to ground states by transferring their energy to Ne atoms through collision there by exciting Ne atoms to energy levels E4 (18.70 eV) and E6 (20.66 eV). This becomes possible because two colliding atoms have identical energy states.

Image…….

The energy difference of 0.05eV is provided by K.E. acquired by He ions. This transfer of energy from He atoms to Ne atoms take place because E6 and E4 energy levels have nearly same values as that of He atoms. Ne atoms do not get excited by directly absorbing energy from pump since they are heavier than He ions. Ne atoms are the active centers and the role of He atoms is to excite Ne atoms which causes population inversion. To ensure proper energy transfer from He to Ne atoms; the proportion of He and Ne atoms is chosen as 10 : 1. This also reduces probability of reverse transfer of energy from Ne atoms to He atoms. The excited states of helium atoms are metastable states hence if the path for their decay is not provided they tend to accumulate and subsequently no He ions are available for excitation of Ne atoms. Decay path for He ions is provided by making the tube narrower so that the excited He atoms collide with walls of tube and return to their ground states.

The E6 and E4 energy states of Ne atoms are metastable states, therefore Ne atoms accumulate in these two states and population inversion between is achieved between E6 - E5, E3 and E4 -E3 levels.
Therefore three LASER transition takes place which are:

  1. E6 to E3 transition: giving red color LASER beam of wavelength 6328 Å
  2. E6 to E5 transition: giving LASER beam in IR region with wavelength 33900 Å
  3. E4 to E3 transition: giving LASER beam in IR region with wavelength 11500 Å

From terminal levels E5 and E3, Ne atoms make spontaneous emission and shows downward transition to E2 level. As E2 is meta-stable state and therefore Ne atoms again accumulate at this level. Atoms from E2 state are required to be brought to ground state E1 because if they accommodate in E2 state, number of atoms in ground state will decrease fast it becomes difficult to maintain population inversion continuously.

In order to avoid this, discharge tube is made narrow due to which probability of collision of atoms with walls increases and atoms can give up excess energy and can return back to ground state easily. This LASER is used as a monochromatic source in interferometry, LASER printers, bar code reading etc. They are also used as reference beam in surveying, alignment in laying pipes etc.

(b) Derive the condition for a thin transparent film of constant thickness to appear bright mid dark when viewed in reflected fight. [7M]

When a monochromatic light beam is incident on a transparent parallel thin film of uniform
thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern.
These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially
reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected
from lower surface of the film along MD and finally emerges along MK
These rays BC and DE interfere and interference fringes are produced. The intensity at any Point
depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and
Determination of path difference between the rays BM and MD. As shown in fig. 1

The geometric path difference between ray 1 and ray 2 = MF + FD - BH

Optical path difference = Δ

\begin{array}{l}=\mu(BM+FD)-\mu BH\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array} \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\Further,\end{array} \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}

But MG=t

BM=\frac t{\cos\;r}\;……….\left(2\right)

Similarly

MD=\frac t{\cos\;r}\;……….\left(3\right)

Now, In ΔBHD

\begin{array}{l}\sin\;i=\frac{BH}{BD}\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\BD=BG+GD\;\;………\left(5\right)\end{array}

But
Therefore, In ΔBGM

\begin{array}{l}\tan\;r=\frac{BG}{MG}\\BG=MG\;\tan\;r\\BG=t\;\tan\;r\end{array}

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

\begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\BD=2t\;\tan\;r\end{array}

Therefore, equation 4 becomes,

BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)

Put equation 2,3 and 6 in equation 1 we get

\begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}

But

\begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\sin\;i=\;\mu\;\sin\;r\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\triangle=2\;\mu t\;\cos\;r\end{array}

When light is reflected from the surface of an optically denser medium, a phase change π is
introduced. Correspondingly, path difference λ/2 is introduced, therefore
The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:

  1. When path difference ∆ = n λ, where n = 0,1,2,………
    Constructive interference takes place and film appears bright.
    Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
    2μ t cos r = (2n +1) λ/2
  2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
    Destructive interference takes place and film appears dark.
    Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
    2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore,
2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.

  1. (a) Calculate the maximum order of diffraction maxima seen from a plane diffraction grating having 5500 lines per cm if light of wavelength 5896 Å falls normally on it. [5M]

We have,

\begin{array}{l}\left(a+b\right)\;\sin\theta=n\lambda\\n=\left(a+b\right)\;\sin\theta\lambda\end{array}

For maximum value of n, {\left[\sin\theta\right]}_{max}=1

\begin{array}{l}n=\frac{a+b}\lambda\\\;\;\;=\frac{\displaystyle\frac{1\;cm}{5500}}{5.893\times10^{-5}cm}\\\;\;\;=\frac1{3.24\times10^{-1}}\\\;\;\;=3.08\\3.08\approx3\;orders\end{array}

(b) Derive Schrodinger's time-independent wave equation. [5M]

Schrodinger’s wave equation determines the motion of atomic particle. Consider a microparticle in motion which is associated with wave function Ψ. The Ψ represents the field of the wave, say. The classical wave equation is of the form,

\frac{\partial^2y}{\partial x^2}=\frac1{\nu^2}\frac{\displaystyle\partial^2y}{\displaystyle\partial t^2}

A solution of this equation is, y\left(x,\;t\right)=Ae^{i\left(kx-wt\right)}

For a microparticle the w and k can be replaced with E and p using Einstein and de-Broglie relations as,

E=\hslash\omega\;and\;p=\hslash k,

Also replacing y(x, t) by \psi(x, t) we may write

\psi\left(x,\;t\right)=Ae^\frac{-\left(Et-px\right)}ℏ

Differentiating with respect to t, we get,

\frac{\partial\psi}{\partial t}=-\frac{i}{\hslash}E\psi

Differentiating twice with respect to x, we get,

\frac{\partial^2\psi}{\partial x^2}=-\frac{p^{2}}{\hslash}\psi

For a free particle we have

E=\frac{p^2}{2m}

and in case of a particle moving in force field characterized by potential energy V, we have

\frac{p^2}{2m}=E-V.

Multiplying above equation by \psi,

\frac{p^2}{2m}\psi=E\psi-V\psi

Substituting for E\psi\;and\;p^2\psi, and  rearranging we get

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=i\hslash\frac{\partial \psi }{\partial t}

Which is time dependent Schrodinger's wave equation, where i\hslash\frac\partial{\partial t}=E, the energy operator.

Therefore,

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=E\psi

This is time independent Schrodinger's wave equation.

(c) Define the term superconductivity. Show that in the superconducting state the material is perfectly diamagnetic. [5M]

Superconductivity: The sudden disappearance of electrical resistance in materials below a certain temperature is known as superconductivity. The materials that exhibit superconductivity and which are in the superconducting state are called superconductors. The temperature at which a normal material turns into a superconductor is called as critical temperature, Tc. Every superconductor has it’s own critical temperature at which it passes over into superconducting state. The superconducting transition is sharp for chemically pure and structurally pure specimen but broad for impure specimens and with structural defects. In 1933 Meissener and Oschsenfield discovered that a superconductor completely expels the magnetic field lines that were initially penetrating it in it’s normal state. This property is independent of the path by which the superconducting state is reached, as shown in the figure. Suppose that magnetic field is applied first to the sample in normal state. The sample is them cooled below Tc, in the presence of magnetic field. It is expected that the magnetic flux through the sample remains unchanged. But, Meissener and Oschsenfield found that magnetic flux was totally expelled by the sample as it becomes superconducting. The expulsion of the magnetic field during the transition from the normal to the superconducting state is called as Meissener effect.

The effect is reversible. When temperature is raised from below Tc, the flux suddenly starts penetrating the specimen at T=Tc as a result of which the material returns back to the normal state. The magnetic induction inside the specimen is given by, B=\mu_0\left(H+M\right)

Where, H is the external magnetic field and M is the magnetization produced in the specimen. At, T<T_c,\;B=0\;and\;\mu_0\left(H+M\right)=0 superconducting state.

Therefore, M = -H

The susceptibility of the material is. x = M/H = -1

Thus, the superconducting state is characterized by perfect diamagnetism. Meissener effect shows that in superconductor not only dB/dt =0 but also B=0. The superconducting state is a characteristic thermodynamic phase of a substance in which material cannot sustain steady electric and magnetic field. Meissener effect is the standard test which conclusively proves that the material is in superconducting state or not. Because of diamagnetism superconducting material strongly repel the external magnets. It leads to levitation effect in which a magnet hovers over superconducting material and also suspension effect in which a chip of superconducting material hangs beneath a magnet.

  1. (a) A slit of width 0.3 mm is illuminated by a light of wavelength 5890 Å. A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern. Calculate the distance between first dark and the next bright fringe form the axis. [5M]

The distance of first dark band from the centre is λD/a and next bright band is 3λD/2a.
Thus, the distance between first dark and next bright will be,

\begin{array}{l}x=\frac{\lambda D}{2a}\\\;\;\;=\frac{5.89\times10^{-5}cm\times40cm}{2\times0.03cm}\\\;\;\;=0.0392\;cm\end{array}

(Note: the focal length f is nearly equal to D, when lens is placed close to the slits.)

(b) An electron is accelerated through 1000 volts and is reflected from a crystal. The first order reflection occurs when glancing angle is 70°. Calculate the interplanar spacing of a crystal. [5M]

We have,

2d\;\sin\theta=n\lambda

Where \begin{array}{l}\lambda=\frac{12.26}{\sqrt v}\overset\circ A\\\;\;\;=\frac{12.26}{\sqrt{1000}}\\\;\;\;=0.3876\;\overset\circ A\end{array}

Thus, \begin{array}{l}d=\frac{n\lambda}{2\;\sin\theta}\\\;\;\;=\frac{1\times0.3876\;\overset\circ A}{2\;\sin\;70}\\\;\;\;=0.2062\;\overset\circ A\end{array}

(c) Explain construction and working of Atomic Force Microscope. [5M]

Atomic force microscopy (AFM) or scanning force microscopy (SFM) is a very high-resolution
type of scanning probe microscopy, with demonstrated resolution on the order of fractions of a nanometer, more than 1000 times better than the optical diffraction limit. The AFM is one of the foremost  tools for imaging, measuring, and manipulating matter at the nanoscale.

The information is gathered by "feeling" the surface with a mechanical probe. Piezoelectric
elements that facilitate tiny but accurate and precise movements on (electronic) command enable the very precise scanning. In some variations, electric potentials can also be scanned using conducting cantilevers. In newer more advanced versions, currents can even be passed through the tip to probe the electrical conductivity or transport of the underlying surface, but this is much more challenging with very few research groups reporting reliable data.

The AFM consists of a cantilever with a sharp tip (probe) at its end that is used to scan the specimen surface. The cantilever is typically silicon or silicon nitride with a tip radius of curvature on the order of nanometers. When the tip is brought into proximity of a sample surface, forces between the tip and the sample lead to a deflection of the cantilever according to Hooke's law.

Depending on the situation, forces that are measured in AFM include mechanical contact force, van der Waals forces, capillary forces, chemical bonding, electrostatic forces, magnetic forces (see magnetic force microscope, MFM), Casimir forces, solvation forces, etc. Along with force, additional quantities may simultaneously be measured through the use of specialized types of probe.

If the tip was scanned at a constant height, a risk would exist that the tip collides with the surface, causing damage. Hence, in most cases a feedback mechanism is employed to adjust the tip-to-sample distance to maintain a constant force between the tip and the sample. Traditionally, the sample is mounted on a piezoelectric tube that can move the sample in the z direction for maintaining a constant force, and the x and y directions for scanning the sample.

Alternatively a 'tripod' configuration of three piezo crystals may be employed, with each
responsible for scanning in the x,y and z directions. This eliminates some of the distortion effects seen with a tube scanner. In newer designs, the tip is mounted on a vertical piezo scanner while the sample is being scanned in X and Y using another piezo block. The resulting map of the area z = f(x, y) represents the topography of the sample.

The AFM can be operated in a number of modes, depending on the application. In general,
possible imaging modes are divided into static (also called contact) modes and a variety of dynamic (or non-contact) modes where the cantilever is vibrated.

Image……

  1. (a) State Heisenber's uncertainty principle. Show that electron cannot pre-exist in free state in a nucleus. [5M]

Statement: “It is impossible to determine exact position and exact momentum both of a particle simultaneously with unlimited accuracy.”

It means that, if the position of the particle is determined accurately then there is uncertainty in the determination of momentum and if the momentum of the particle is determined accurately then there is uncertainty in the determination of its position.

Mathematically, “The product of uncertainties in simultaneous measurement of position and momentum of a microscopic particle is always of the order of Planck’s constant or greater than or equal to h/2π”.
Thus mathematically,

Δx Δpx ≥ ђ [ ђ = h/2 π ]

Or

Δx Δpx ≥ h/2π

This is called as Heisenberg’s uncertainty relation. Where, Δx = Uncertainty in measurement of position, ΔpX = uncertainty in measurement of momentum.
If Δx is small i.e. position is determined more accurately then ΔpX will be large i.e. momentum is determined less accurately and vice-versa.
If electron is inside the nucleus, we will get it within its diameter. i.e. \triangle x=2\times10^{-14}.
Now, Δx Δp = h/2π or Δp ≥ h/2π Δx ≈ 5 x 10-21 kgm/s.

This, minimum energy,

\begin{array}{l}E=\;p.c\;=\;mc.c\;\;\\\;\;\;=\;3.2\;x\;10-12\;J\;\\\;\;\;=\;20\;MeV.\end{array}

Thus if electron is inside nucleus its energy must be about 20 MeV. But, the maximum energy of electron found is 3 MeV, hence electron cannot reside the nucleus.

(b) Draw a labelled diagram and explain construction and working of CRT. [5M]

Cathode Ray Tube is an important part of CRO and is said to be the heart of CRO. It generates the electron beam, focuses it and accelerates it towards the fluorescent screen. It mainly consists of (i)Electron gun assembly, (ii) Deflection plate assembly, and (iii) Fluorescent screen as shown in figure.

1 Electron gun assembly
The function of electron gun is to produce a sharply focussed beam of electrons, which are accelerated to a high velocity. It consists of a heater, cathode, control grid, pre accelerating anode, focusing anode and an accelerating anode.

The cathode is in the form of a cylinder. The electrons emitted from this indirectly heated cathode come out through a small hole and enter control grid. The control grid is negatively biased cylinder with a centrally located hole. It controls the number of electrons reaching the anode and thus the intensity of electron beam is controlled by the grid. These electrons are accelerated by a high positive potential to preaccelerating and accelerating anodes

Image…….

2 Deflection plate assembly:
It consists of two pairs of deflecting plates for deflecting the beam of electron both in the vertical and horizontal directions. One pair of plates, called the X-plates, is mounted vertically and it deflects the electron beam in the horizontal direction due to the electric field produced in horizontal plane by applying certain potential across these plates. Another pair of plates called the Y-plates is mounted horizontally which produces the electric field in a vertical plane, after applying a potential difference across them. This produces vertical deflection of the electron beam. The plates are flared so as to allow the beam to pass
through them without striking the plates.

3 Fluorescent Screen
The screen of CRT is coated with a fluorescent material like Zink orthosilicate, in general called as phosphor. It absorbs the kinetic energy of electrons striking it and re-emits in the form of light. Hence the screen is called as fluorescent screen. The colour of the trace on the screen depends on the type of material.

All these parts of CRT are enclosed in a funnel shaped highly evacuated glass envelope. The
inner surface of the flared part of this envelope is coated with an aqueous solution of graphite called aquadag, which is internally connected to the accelerating anode. This layer performs two functions-
(1) It accelerates the electron beam, (2) It collects the secondary electrons produced by secondary emission when electron beam strikes the screen.

(c) Exploits top down and bottom up approaches to prepare nanomaterials. [5M]

Engineered nano-objects and nanomaterials intended for industrial use may be synthesized using two different approaches known as the op down Approach and the Bottom Up Approach.

  1. Top-down approach
    The most common top-down approach to fabrication involves lithographic patterning techniques using short-wavelength optical sources. A key advantage of the top-down approach as developed in the fabrication of integrated circuits is that the parts are both patterned and built in place, so that no assembly step is needed. Optical lithography is a relatively mature field because of the high degree of refinement in microelectronic chip manufacturing, with current short-wavelength optical lithography techniques reaching dimensions just below 100 nanometers (the traditional threshold definition of the nanoscale).

Shorter-wavelength sources, such as extreme ultraviolet and X-ray, are being developed to allow lithographic printing techniques to reach dimensions from 10 to 100 nanometers. Scanning beam techniques such as electron-beam lithography provide patterns down to about 20 nanometers. Here the pattern is written by sweeping a finely focused electron beam across the surface. Focused ion beams are also used for direct processing and patterning of wafers, although with somewhat less resolution than in electron-beam lithography. Still-smaller features are obtained by using scanning probes to deposit or remove thin layers. Mechanical printing techniques—nanoscale imprinting, stamping, and molding—have been extended to the surprisingly small dimensions of about 20 to 40 nanometers.

  1. Bottom-up approach
    Bottom-up or self-assembly approach to nanofabrication use chemical or physical forces operating at the nanoscale to assemble basic units into larger structures. As component size decreases in nanofabrication, bottom-up approaches provide an increasingly important complement to top-down techniques. Inspiration for bottom-up approach comes from biological systems, where nature has harnessed chemical forces to create essentially all the structures needed by life. Researchers hope to replicate nature‘s ability to produce small clusters of specific atoms, which can then self-assemble into more-elaborate structures. A number of bottom-up approaches have been developed for producing nanoparticles, ranging from condensation of atomic vapors on surfaces to coalescence of atoms in liquids. For example, liquid-phase techniques have been developed to produce size-selected nanoparticles of semiconductor, magnetic, and other materials.

An example of self-assembly that achieves a limited degree of control over both formation and organization is the growth of quantum dots. Indium gallium arsenide (InGaAs) dots can be formed by growing thin layers of InGaAs on GaAs in such a manner that repulsive forces caused by compressive strain in the InGaAs layer results in the formation of isolated quantum dots. After the growth of multiple layer pairs, a fairly uniform spacing of the dots can be achieved. Another example of self-assembly of an intricate structure is the formation of carbon nanotubes under the right set of chemical and temperature conditions.

Image…….B.E. Second Semester All Branches (C.B.S.)
Time: 2 hours
Maximum marks: 60

Notes :
1.Question No. 1 is compulsory.
2. Attempt any three questions from Q.2 to Q.6.

  1. Used suitable data wherever required.
  2. Figures to the right indicate full mark.
  3. Attempt any five of the following :- [15M]
    (a) Comment on colours in a soap film in sunlight.

If sunlight is incident on a soap film, the optical path difference will vary from one color to other due to change of wavelength with color. Therefore the film will appear to be colored and the color seen will be of that rays for which constructive interference occurs. If the sunlight is not parallel then the optical path difference will change due to change in angle of incidence. Hence, the film will show different colors when viewed from different direction.

(b) What is Rayleigh's criterion of resolution ? Define resolving power of a grating.

When an object is seen through a optical instrument, diffraction occurs as the light passes through the aperture. Hence in actual practice the diffraction pattern of the object is seen when viewed through any optical instrument. Because of which there is a limit on the ability of that instrument to view the two closely spaced point objects separately.

According to Rayleigh’s criterion the closely spaced point objects are said to be just resolved, or can be seen separately, when the principle maxima of one objects fall upon the first minimum of second object and vice versa. If the distance decreases then they are not resolved and if distance increased then they are said to be well resolved.

Resolving power of grating can be defined as the reciprocal of angular or linear distance for which two spectral lines can be seen separately and is given by

R=λ/Δλ.

(c) Calculate V number for an optical fiber having numerical aperture 0.25 and core diameter 20 μm if it is operated at 1.55 μm.

Given,
NA=0.25, d= 20 x 10-6 m, λo= 1.55 x 10-6 m,

We have,

\begin{array}{l}V=\frac{\pi d}\lambda\left[NA\right]\\\;\;\;=\frac{3.14\times20\times10^{-6}}{1.55\times10^{-6}}\times0.25\\\;\;\;=10.129\end{array}

(d) Compare light from ordinary source with laser light.

1) Light from ordinary source is not directional, whereas laser light is highly directional.

2) Ordinary light is poly-chromatic whereas LASER is highly monochromatic and contains a very
narrow range of the order of a few Angstroms (<10 Å ). 3) The intensity of ordinary light is very small whereas LASER emits light in the form of a narrow beam with its energy concentrated in a small region of space .Therefore, the beam intensity would be tremendously large and stays constant with distance since it travels in the form of plane waves. 4) Ordinary light is highly divergent whereas Light from a LASER propagates in the form of plane waves and it has very small divergence. 5) ordinary light is incoherent light since they emit light waves of random wavelengths having random phases. While, the waves emitted by LASER source are in phase and having same frequency. Therefore, light emitted by LASER is highly coherent. (e) How phase difference between two signals is measured using CRO? When two sine waves oscillating in mutually perpendicular direction are of same frequency, the Lissajous pattern takes the form of an ellipse as shown in the figure. This ellipse is used to determine the phase difference between two signals, by using the formula \phi=\sin^{-1}\left(\frac{AA'}{BB'}\right) By measuring the length AA’ and BB’ on the oscillogram and substituting the values in the above formula phase difference can be determined. Images……. (f) What are the properties of matter waves? 1. Wavelength of Matter waves depends upon mass and velocity of particle \left(\lambda=\frac h{m\nu}\right), where as wavelength of EM waves depends upon energy \left(\lambda=\frac{hc}E\right). 2. Velocity of matter wave depend upon velocity of particle and greater than velocity of light, whereas velocity of EM wave is constant equal to velocity of light. 3. Matter waves are not real waves but are probability waves only, whereas EM waves are real waves with periodic variation of E and B vector in them. 4. Matter waves are produced by a motion of particle and are independent of charges, whereas EM waves are produced by motion of charged particle only. (g) A superconductor has a critical temperature 3.7°K at zero magnetic field. At O°K the critical magnetic field is 0.0306 Tesla. What is the critical magnetic field at temperature 2.0°K? We have, \begin{array}{l}H_c\left(T\right)=H_c\left(o\right)\left[1-\left(\frac T{T_c}\right)\right]^2\\\;\;\;\;\;\;\;\;\;\;=0.0306\left[1-\left(\frac2{3.7}\right)\right]^2\\H_c\left(T\right)=0.00646\;K\\\end{array} 2. (a) Show that the diameter of Newton's nth dark ring is proportional to square root of ring number. In Newton's rings experiment the diameter of 5th dark ring was 0.336 cm and that of 15th dark ring was 0.590 cm. Calculate the radius of curvature of plano-convex lens if wavelength of light used is 5890 Å. [8M] Let R be the radius of curvature of the lens. Let a dark ring be located at the point B. The thickness of the air film at B is t and AE = r, the radius of the ring at thickness t. By using the Pythagoras theorem to triangle AEO we get, AO2 =AE2+EO2 R2 = r2+ (R-t)2 = r2 + R2 – 2Rt + t2 r2 = 2Rt or t = r2/2R  ,As t is very small, t2 can be neglected Using relation 2t = nλ, for dark rings. r2 = nλR r=\sqrt{n\lambda R} Dark ring. The Diameter of dark ring is given by D_n=\sqrt{4n\lambda R} Since, λ and  R are constants D_n\alpha\sqrt n i.e. the diameter of the Newton's nth dark ring is directly proportional to square root of the ring number. we have,

\begin{array}{l}R=\frac{D_n^2}{4n\lambda}\\\;\;\;=\frac{\left(0.336\times10^{-2}\right)^2}{4\times5\times5.89\times10^{-7}}\\\;\;\;=0.000958\times10^3\\\;\;\;=95.8\times10^{-2}m\\\;\;\;=95.8\;cm\end{array}<a href="b">/latex</a> Derive an expression for numerical aperture of step index optical fiber. What are the advantages of using an optical fiber? [7M] The main function of the optical fiber is to accept and transmit as much as light from the source as possible. The light gathering ability of the fiber depends on two factors, namely core size and numerical aperture. The numerical aperture determined by the acceptance angle and fractional refractive index change. Acceptance angle and acceptance cone image…… Let us consider an optical fiber into which light is launched. Let refractive index of core be µ1 and refractive index of cladding be µ2.The refractive index of cladding is less than refractive index of core. Let µ0 be the refractive index of the medium from which light is launched into the fiber. Let a light ray enter the fiber at an angle θi to the axis of the fiber. The rays refracts at an angle θr and strike the core cladding interface at an angle [latex] \phi

. If \phi is greater than the critical angle \phic, the rays undergoes total internal reflection at the interface, since µ1>µ2 as long as the angle \phi is greater than critical angle \phic, the light will stay within the fiber as shown figure.

Let us now compute the incident angle θi for which  \phi\;\geq\;\phi_c such that light rebounds within the
fiber. Therefore From Snell’s law,

\begin{array}{l}\frac{\sin\theta_i}{\sin\theta_r}=\frac{\mu_1}{\mu_2}\\\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_2}\sin\theta_r-----\left(i\right)\end{array}

If θi is increased beyond a limit, \phi will drop below the critical value \phic and the ray escape from the sides walls of the fiber.

\sin\theta_r=\sin\left(90-\phi\right)=\cos\phi\;-----\left(ii\right)

Using equation 1 and 2

\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi

when \phi=\phi_c

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi_c-----\left(iii\right)

But by the condition of total internal reflection

\sin\phi_c=\frac{\displaystyle\mu_2}{\displaystyle\mu_1}

Now, Squaring both side

\begin{array}{l}\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\;;\;1-\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\\\cos^2\phi_c=\frac{\mu_1^2-\mu_2^2}{\mu_1^2}\;;\;\cos^2\phi_c=\frac{\displaystyle\mu_1^2-\mu_2^2}{\displaystyle\mu_1^2}\\\cos\phi_c=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_1}-----\left(iv\right)\end{array}

Put equation (iv) in equation (3), we get,

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_0}

Incident ray is launched from air medium for which, \mu_0=1 and substituting \theta_{i\left(max\right)}=\theta_0

we get,

\begin{array}{l}\sin\theta_0=\sqrt{\mu_1^2-\mu_2^2}\\\theta_0=\sin^{-1}=\sqrt{\mu_1^2-\mu_2^2}\end{array}
The angle θ0 is called the acceptance angle of the fiber.

Acceptance angle: It is defined as the maximum angle that a light ray can have relative to the axis of the fiber and propagate down the fiber.

Acceptance Cone: The light ray contained within the cone having a full angle 2θ0 are accepted and transmitted along the fiber. Therefore the cone is called the acceptance cone.

Fractional refractive index change
The fractional difference Δ between the refractive indices of the core and the cladding is known as Fractional refractive index change.

\triangle=\frac{\displaystyle\mu_1-\mu_2}{\mu_1}

Numerical aperture: The numerical aperture (NA) is defined as the acceptance angle.

Thus

\begin{array}{l}N.A.=\sin\theta_0\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}{2\mu_1}}2\mu_1\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1-\mu_2\right)}{\mu_1}}2\mu_1^2\;\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1+\mu_2}2=\mu_1\right)\\N.A.=\mu_1\sqrt{2\triangle}---\left(i\right)\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1-\mu_2}{\mu_1}=\triangle\right)\end{array}

Numerical aperture determines the light gathering ability of the fiber. It is measure of the amount of the light that can be accepted by fiber. It is seen from equation (i) that NA. dependent only on the refractive indices of the core and cladding materials. Its value range from 0.13 to 0.50. A large NA implies that a fiber will accept large amount of light from the source.

  1. (a) Explain construction and working of He-Ne laser. What are its merits ? [8M]

Construction: It consists of a glass discharge tube of about 30 cm long and 1 cm diameter. It is filled with mixture of He and Ne gas in the ratio of 10:1. Two electrodes are provided to produce discharge in the gas and connected to high voltage power supply. The tube is hermetically sealed by inclined windows arranged at its two ends. On the axis of the tube two reflectors are fixed which constitute optical resonator cavity. The distance between the mirrors is adjusted such that it supports standing wave pattern.

Image…..

Working:
The energy level diagram is shown in figure. It employs four level pumping scheme. When the power supply is put on; due to electric field the gas is ionized producing electrons and ions of gas. The electrons and ions are accelerated towards anode and cathode. The electrons transfer their energy to He ions due to collision and are excited to higher state which lie at 19.81 eV and 20.16 eV above the ground state. He- atoms are lighter and hence readily excited. These levels E2 and E3 are metastable states and hence the excited He atoms do not return to ground states through spontaneous emission process. However, the He atoms return to ground states by transferring their energy to Ne atoms through collision there by exciting Ne atoms to energy levels E4 (18.70 eV) and E6 (20.66 eV). This becomes possible because two colliding atoms have identical energy states.

Image…….

The energy difference of 0.05eV is provided by K.E. acquired by He ions. This transfer of energy from He atoms to Ne atoms take place because E6 and E4 energy levels have nearly same values as that of He atoms. Ne atoms do not get excited by directly absorbing energy from pump since they are heavier than He ions. Ne atoms are the active centers and the role of He atoms is to excite Ne atoms which causes population inversion. To ensure proper energy transfer from He to Ne atoms; the proportion of He and Ne atoms is chosen as 10 : 1. This also reduces probability of reverse transfer of energy from Ne atoms to He atoms. The excited states of helium atoms are metastable states hence if the path for their decay is not provided they tend to accumulate and subsequently no He ions are available for excitation of Ne atoms. Decay path for He ions is provided by making the tube narrower so that the excited He atoms collide with walls of tube and return to their ground states.

The E6 and E4 energy states of Ne atoms are metastable states, therefore Ne atoms accumulate in these two states and population inversion between is achieved between E6 - E5, E3 and E4 -E3 levels.
Therefore three LASER transition takes place which are:

  1. E6 to E3 transition: giving red color LASER beam of wavelength 6328 Å
  2. E6 to E5 transition: giving LASER beam in IR region with wavelength 33900 Å
  3. E4 to E3 transition: giving LASER beam in IR region with wavelength 11500 Å

From terminal levels E5 and E3, Ne atoms make spontaneous emission and shows downward transition to E2 level. As E2 is meta-stable state and therefore Ne atoms again accumulate at this level. Atoms from E2 state are required to be brought to ground state E1 because if they accommodate in E2 state, number of atoms in ground state will decrease fast it becomes difficult to maintain population inversion continuously.

In order to avoid this, discharge tube is made narrow due to which probability of collision of atoms with walls increases and atoms can give up excess energy and can return back to ground state easily. This LASER is used as a monochromatic source in interferometry, LASER printers, bar code reading etc. They are also used as reference beam in surveying, alignment in laying pipes etc.

(b) Derive the condition for a thin transparent film of constant thickness to appear bright mid dark when viewed in reflected fight. [7M]

When a monochromatic light beam is incident on a transparent parallel thin film of uniform
thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern.
These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially
reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected
from lower surface of the film along MD and finally emerges along MK
These rays BC and DE interfere and interference fringes are produced. The intensity at any Point
depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and
Determination of path difference between the rays BM and MD. As shown in fig. 1

The geometric path difference between ray 1 and ray 2 = MF + FD - BH

Optical path difference = Δ

\begin{array}{l}=\mu(BM+FD)-\mu BH\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array} \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\Further,\end{array} \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}

But MG=t

BM=\frac t{\cos\;r}\;……….\left(2\right)

Similarly

MD=\frac t{\cos\;r}\;……….\left(3\right)

Now, In ΔBHD

\begin{array}{l}\sin\;i=\frac{BH}{BD}\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\BD=BG+GD\;\;………\left(5\right)\end{array}

But
Therefore, In ΔBGM

\begin{array}{l}\tan\;r=\frac{BG}{MG}\\BG=MG\;\tan\;r\\BG=t\;\tan\;r\end{array}

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

\begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\BD=2t\;\tan\;r\end{array}

Therefore, equation 4 becomes,

BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)

Put equation 2,3 and 6 in equation 1 we get

\begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}

But

\begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\sin\;i=\;\mu\;\sin\;r\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\triangle=2\;\mu t\;\cos\;r\end{array}

When light is reflected from the surface of an optically denser medium, a phase change π is
introduced. Correspondingly, path difference λ/2 is introduced, therefore
The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:

  1. When path difference ∆ = n λ, where n = 0,1,2,………
    Constructive interference takes place and film appears bright.
    Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
    2μ t cos r = (2n +1) λ/2
  2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
    Destructive interference takes place and film appears dark.
    Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
    2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore,
2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.

  1. (a) Calculate the maximum order of diffraction maxima seen from a plane diffraction grating having 5500 lines per cm if light of wavelength 5896 Å falls normally on it. [5M]

We have,

\begin{array}{l}\left(a+b\right)\;\sin\theta=n\lambda\\n=\left(a+b\right)\;\sin\theta\lambda\end{array}

For maximum value of n, {\left[\sin\theta\right]}_{max}=1

\begin{array}{l}n=\frac{a+b}\lambda\\\;\;\;=\frac{\displaystyle\frac{1\;cm}{5500}}{5.893\times10^{-5}cm}\\\;\;\;=\frac1{3.24\times10^{-1}}\\\;\;\;=3.08\\3.08\approx3\;orders\end{array}

(b) Derive Schrodinger's time-independent wave equation. [5M]

Schrodinger’s wave equation determines the motion of atomic particle. Consider a microparticle in motion which is associated with wave function Ψ. The Ψ represents the field of the wave, say. The classical wave equation is of the form,

\frac{\partial^2y}{\partial x^2}=\frac1{\nu^2}\frac{\displaystyle\partial^2y}{\displaystyle\partial t^2}

A solution of this equation is, y\left(x,\;t\right)=Ae^{i\left(kx-wt\right)}

For a microparticle the w and k can be replaced with E and p using Einstein and de-Broglie relations as,

E=\hslash\omega\;and\;p=\hslash k,

Also replacing y(x, t) by \psi(x, t) we may write

\psi\left(x,\;t\right)=Ae^\frac{-\left(Et-px\right)}ℏ

Differentiating with respect to t, we get,

\frac{\partial\psi}{\partial t}=-\frac{i}{\hslash}E\psi

Differentiating twice with respect to x, we get,

\frac{\partial^2\psi}{\partial x^2}=-\frac{p^{2}}{\hslash}\psi

For a free particle we have

E=\frac{p^2}{2m}

and in case of a particle moving in force field characterized by potential energy V, we have

\frac{p^2}{2m}=E-V.

Multiplying above equation by \psi,

\frac{p^2}{2m}\psi=E\psi-V\psi

Substituting for E\psi\;and\;p^2\psi, and  rearranging we get

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=i\hslash\frac{\partial \psi }{\partial t}

Which is time dependent Schrodinger's wave equation, where i\hslash\frac\partial{\partial t}=E, the energy operator.

Therefore,

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=E\psi

This is time independent Schrodinger's wave equation.

(c) Define the term superconductivity. Show that in the superconducting state the material is perfectly diamagnetic. [5M]

Superconductivity: The sudden disappearance of electrical resistance in materials below a certain temperature is known as superconductivity. The materials that exhibit superconductivity and which are in the superconducting state are called superconductors. The temperature at which a normal material turns into a superconductor is called as critical temperature, Tc. Every superconductor has it’s own critical temperature at which it passes over into superconducting state. The superconducting transition is sharp for chemically pure and structurally pure specimen but broad for impure specimens and with structural defects. In 1933 Meissener and Oschsenfield discovered that a superconductor completely expels the magnetic field lines that were initially penetrating it in it’s normal state. This property is independent of the path by which the superconducting state is reached, as shown in the figure. Suppose that magnetic field is applied first to the sample in normal state. The sample is them cooled below Tc, in the presence of magnetic field. It is expected that the magnetic flux through the sample remains unchanged. But, Meissener and Oschsenfield found that magnetic flux was totally expelled by the sample as it becomes superconducting. The expulsion of the magnetic field during the transition from the normal to the superconducting state is called as Meissener effect.

The effect is reversible. When temperature is raised from below Tc, the flux suddenly starts penetrating the specimen at T=Tc as a result of which the material returns back to the normal state. The magnetic induction inside the specimen is given by, B=\mu_0\left(H+M\right)

Where, H is the external magnetic field and M is the magnetization produced in the specimen. At, T<T_c,\;B=0\;and\;\mu_0\left(H+M\right)=0 superconducting state.

Therefore, M = -H

The susceptibility of the material is. x = M/H = -1

Thus, the superconducting state is characterized by perfect diamagnetism. Meissener effect shows that in superconductor not only dB/dt =0 but also B=0. The superconducting state is a characteristic thermodynamic phase of a substance in which material cannot sustain steady electric and magnetic field. Meissener effect is the standard test which conclusively proves that the material is in superconducting state or not. Because of diamagnetism superconducting material strongly repel the external magnets. It leads to levitation effect in which a magnet hovers over superconducting material and also suspension effect in which a chip of superconducting material hangs beneath a magnet.

  1. (a) A slit of width 0.3 mm is illuminated by a light of wavelength 5890 Å. A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern. Calculate the distance between first dark and the next bright fringe form the axis. [5M]

The distance of first dark band from the centre is λD/a and next bright band is 3λD/2a.
Thus, the distance between first dark and next bright will be,

\begin{array}{l}x=\frac{\lambda D}{2a}\\\;\;\;=\frac{5.89\times10^{-5}cm\times40cm}{2\times0.03cm}\\\;\;\;=0.0392\;cm\end{array}

(Note: the focal length f is nearly equal to D, when lens is placed close to the slits.)

(b) An electron is accelerated through 1000 volts and is reflected from a crystal. The first order reflection occurs when glancing angle is 70°. Calculate the interplanar spacing of a crystal. [5M]

We have,

2d\;\sin\theta=n\lambda

Where \begin{array}{l}\lambda=\frac{12.26}{\sqrt v}\overset\circ A\\\;\;\;=\frac{12.26}{\sqrt{1000}}\\\;\;\;=0.3876\;\overset\circ A\end{array}

Thus, \begin{array}{l}d=\frac{n\lambda}{2\;\sin\theta}\\\;\;\;=\frac{1\times0.3876\;\overset\circ A}{2\;\sin\;70}\\\;\;\;=0.2062\;\overset\circ A\end{array}

(c) Explain construction and working of Atomic Force Microscope. [5M]

Atomic force microscopy (AFM) or scanning force microscopy (SFM) is a very high-resolution
type of scanning probe microscopy, with demonstrated resolution on the order of fractions of a nanometer, more than 1000 times better than the optical diffraction limit. The AFM is one of the foremost  tools for imaging, measuring, and manipulating matter at the nanoscale.

The information is gathered by "feeling" the surface with a mechanical probe. Piezoelectric
elements that facilitate tiny but accurate and precise movements on (electronic) command enable the very precise scanning. In some variations, electric potentials can also be scanned using conducting cantilevers. In newer more advanced versions, currents can even be passed through the tip to probe the electrical conductivity or transport of the underlying surface, but this is much more challenging with very few research groups reporting reliable data.

The AFM consists of a cantilever with a sharp tip (probe) at its end that is used to scan the specimen surface. The cantilever is typically silicon or silicon nitride with a tip radius of curvature on the order of nanometers. When the tip is brought into proximity of a sample surface, forces between the tip and the sample lead to a deflection of the cantilever according to Hooke's law.

Depending on the situation, forces that are measured in AFM include mechanical contact force, van der Waals forces, capillary forces, chemical bonding, electrostatic forces, magnetic forces (see magnetic force microscope, MFM), Casimir forces, solvation forces, etc. Along with force, additional quantities may simultaneously be measured through the use of specialized types of probe.

If the tip was scanned at a constant height, a risk would exist that the tip collides with the surface, causing damage. Hence, in most cases a feedback mechanism is employed to adjust the tip-to-sample distance to maintain a constant force between the tip and the sample. Traditionally, the sample is mounted on a piezoelectric tube that can move the sample in the z direction for maintaining a constant force, and the x and y directions for scanning the sample.

Alternatively a 'tripod' configuration of three piezo crystals may be employed, with each
responsible for scanning in the x,y and z directions. This eliminates some of the distortion effects seen with a tube scanner. In newer designs, the tip is mounted on a vertical piezo scanner while the sample is being scanned in X and Y using another piezo block. The resulting map of the area z = f(x, y) represents the topography of the sample.

The AFM can be operated in a number of modes, depending on the application. In general,
possible imaging modes are divided into static (also called contact) modes and a variety of dynamic (or non-contact) modes where the cantilever is vibrated.

Image……

  1. (a) State Heisenber's uncertainty principle. Show that electron cannot pre-exist in free state in a nucleus. [5M]

Statement: “It is impossible to determine exact position and exact momentum both of a particle simultaneously with unlimited accuracy.”

It means that, if the position of the particle is determined accurately then there is uncertainty in the determination of momentum and if the momentum of the particle is determined accurately then there is uncertainty in the determination of its position.

Mathematically, “The product of uncertainties in simultaneous measurement of position and momentum of a microscopic particle is always of the order of Planck’s constant or greater than or equal to h/2π”.
Thus mathematically,

Δx Δpx ≥ ђ [ ђ = h/2 π ]

Or

Δx Δpx ≥ h/2π

This is called as Heisenberg’s uncertainty relation. Where, Δx = Uncertainty in measurement of position, ΔpX = uncertainty in measurement of momentum.
If Δx is small i.e. position is determined more accurately then ΔpX will be large i.e. momentum is determined less accurately and vice-versa.
If electron is inside the nucleus, we will get it within its diameter. i.e. \triangle x=2\times10^{-14}.
Now, Δx Δp = h/2π or Δp ≥ h/2π Δx ≈ 5 x 10-21 kgm/s.

This, minimum energy,

\begin{array}{l}E=\;p.c\;=\;mc.c\;\;\\\;\;\;=\;3.2\;x\;10-12\;J\;\\\;\;\;=\;20\;MeV.\end{array}

Thus if electron is inside nucleus its energy must be about 20 MeV. But, the maximum energy of electron found is 3 MeV, hence electron cannot reside the nucleus.

(b) Draw a labelled diagram and explain construction and working of CRT. [5M]

Cathode Ray Tube is an important part of CRO and is said to be the heart of CRO. It generates the electron beam, focuses it and accelerates it towards the fluorescent screen. It mainly consists of (i)Electron gun assembly, (ii) Deflection plate assembly, and (iii) Fluorescent screen as shown in figure.

1 Electron gun assembly
The function of electron gun is to produce a sharply focussed beam of electrons, which are accelerated to a high velocity. It consists of a heater, cathode, control grid, pre accelerating anode, focusing anode and an accelerating anode.

The cathode is in the form of a cylinder. The electrons emitted from this indirectly heated cathode come out through a small hole and enter control grid. The control grid is negatively biased cylinder with a centrally located hole. It controls the number of electrons reaching the anode and thus the intensity of electron beam is controlled by the grid. These electrons are accelerated by a high positive potential to preaccelerating and accelerating anodes

Image…….

2 Deflection plate assembly:
It consists of two pairs of deflecting plates for deflecting the beam of electron both in the vertical and horizontal directions. One pair of plates, called the X-plates, is mounted vertically and it deflects the electron beam in the horizontal direction due to the electric field produced in horizontal plane by applying certain potential across these plates. Another pair of plates called the Y-plates is mounted horizontally which produces the electric field in a vertical plane, after applying a potential difference across them. This produces vertical deflection of the electron beam. The plates are flared so as to allow the beam to pass
through them without striking the plates.

3 Fluorescent Screen
The screen of CRT is coated with a fluorescent material like Zink orthosilicate, in general called as phosphor. It absorbs the kinetic energy of electrons striking it and re-emits in the form of light. Hence the screen is called as fluorescent screen. The colour of the trace on the screen depends on the type of material.

All these parts of CRT are enclosed in a funnel shaped highly evacuated glass envelope. The
inner surface of the flared part of this envelope is coated with an aqueous solution of graphite called aquadag, which is internally connected to the accelerating anode. This layer performs two functions-
(1) It accelerates the electron beam, (2) It collects the secondary electrons produced by secondary emission when electron beam strikes the screen.

(c) Exploits top down and bottom up approaches to prepare nanomaterials. [5M]

Engineered nano-objects and nanomaterials intended for industrial use may be synthesized using two different approaches known as the op down Approach and the Bottom Up Approach.

  1. Top-down approach
    The most common top-down approach to fabrication involves lithographic patterning techniques using short-wavelength optical sources. A key advantage of the top-down approach as developed in the fabrication of integrated circuits is that the parts are both patterned and built in place, so that no assembly step is needed. Optical lithography is a relatively mature field because of the high degree of refinement in microelectronic chip manufacturing, with current short-wavelength optical lithography techniques reaching dimensions just below 100 nanometers (the traditional threshold definition of the nanoscale).

Shorter-wavelength sources, such as extreme ultraviolet and X-ray, are being developed to allow lithographic printing techniques to reach dimensions from 10 to 100 nanometers. Scanning beam techniques such as electron-beam lithography provide patterns down to about 20 nanometers. Here the pattern is written by sweeping a finely focused electron beam across the surface. Focused ion beams are also used for direct processing and patterning of wafers, although with somewhat less resolution than in electron-beam lithography. Still-smaller features are obtained by using scanning probes to deposit or remove thin layers. Mechanical printing techniques—nanoscale imprinting, stamping, and molding—have been extended to the surprisingly small dimensions of about 20 to 40 nanometers.

  1. Bottom-up approach
    Bottom-up or self-assembly approach to nanofabrication use chemical or physical forces operating at the nanoscale to assemble basic units into larger structures. As component size decreases in nanofabrication, bottom-up approaches provide an increasingly important complement to top-down techniques. Inspiration for bottom-up approach comes from biological systems, where nature has harnessed chemical forces to create essentially all the structures needed by life. Researchers hope to replicate nature‘s ability to produce small clusters of specific atoms, which can then self-assemble into more-elaborate structures. A number of bottom-up approaches have been developed for producing nanoparticles, ranging from condensation of atomic vapors on surfaces to coalescence of atoms in liquids. For example, liquid-phase techniques have been developed to produce size-selected nanoparticles of semiconductor, magnetic, and other materials.

An example of self-assembly that achieves a limited degree of control over both formation and organization is the growth of quantum dots. Indium gallium arsenide (InGaAs) dots can be formed by growing thin layers of InGaAs on GaAs in such a manner that repulsive forces caused by compressive strain in the InGaAs layer results in the formation of isolated quantum dots. After the growth of multiple layer pairs, a fairly uniform spacing of the dots can be achieved. Another example of self-assembly of an intricate structure is the formation of carbon nanotubes under the right set of chemical and temperature conditions.

Image…….B.E. Second Semester All Branches (C.B.S.)
Time: 2 hours
Maximum marks: 60

Notes :
1.Question No. 1 is compulsory.
2. Attempt any three questions from Q.2 to Q.6.

  1. Used suitable data wherever required.
  2. Figures to the right indicate full mark.
  3. Attempt any five of the following :- [15M]
    (a) Comment on colours in a soap film in sunlight.

If sunlight is incident on a soap film, the optical path difference will vary from one color to other due to change of wavelength with color. Therefore the film will appear to be colored and the color seen will be of that rays for which constructive interference occurs. If the sunlight is not parallel then the optical path difference will change due to change in angle of incidence. Hence, the film will show different colors when viewed from different direction.

(b) What is Rayleigh's criterion of resolution ? Define resolving power of a grating.

When an object is seen through a optical instrument, diffraction occurs as the light passes through the aperture. Hence in actual practice the diffraction pattern of the object is seen when viewed through any optical instrument. Because of which there is a limit on the ability of that instrument to view the two closely spaced point objects separately.

According to Rayleigh’s criterion the closely spaced point objects are said to be just resolved, or can be seen separately, when the principle maxima of one objects fall upon the first minimum of second object and vice versa. If the distance decreases then they are not resolved and if distance increased then they are said to be well resolved.

Resolving power of grating can be defined as the reciprocal of angular or linear distance for which two spectral lines can be seen separately and is given by

R=λ/Δλ.

(c) Calculate V number for an optical fiber having numerical aperture 0.25 and core diameter 20 μm if it is operated at 1.55 μm.

Given,
NA=0.25, d= 20 x 10-6 m, λo= 1.55 x 10-6 m,

We have,

\begin{array}{l}V=\frac{\pi d}\lambda\left[NA\right]\\\;\;\;=\frac{3.14\times20\times10^{-6}}{1.55\times10^{-6}}\times0.25\\\;\;\;=10.129\end{array}

(d) Compare light from ordinary source with laser light.

1) Light from ordinary source is not directional, whereas laser light is highly directional.

2) Ordinary light is poly-chromatic whereas LASER is highly monochromatic and contains a very
narrow range of the order of a few Angstroms (<10 Å ). 3) The intensity of ordinary light is very small whereas LASER emits light in the form of a narrow beam with its energy concentrated in a small region of space .Therefore, the beam intensity would be tremendously large and stays constant with distance since it travels in the form of plane waves. 4) Ordinary light is highly divergent whereas Light from a LASER propagates in the form of plane waves and it has very small divergence. 5) ordinary light is incoherent light since they emit light waves of random wavelengths having random phases. While, the waves emitted by LASER source are in phase and having same frequency. Therefore, light emitted by LASER is highly coherent. (e) How phase difference between two signals is measured using CRO? When two sine waves oscillating in mutually perpendicular direction are of same frequency, the Lissajous pattern takes the form of an ellipse as shown in the figure. This ellipse is used to determine the phase difference between two signals, by using the formula \phi=\sin^{-1}\left(\frac{AA'}{BB'}\right) By measuring the length AA’ and BB’ on the oscillogram and substituting the values in the above formula phase difference can be determined. Images……. (f) What are the properties of matter waves? 1. Wavelength of Matter waves depends upon mass and velocity of particle \left(\lambda=\frac h{m\nu}\right), where as wavelength of EM waves depends upon energy \left(\lambda=\frac{hc}E\right). 2. Velocity of matter wave depend upon velocity of particle and greater than velocity of light, whereas velocity of EM wave is constant equal to velocity of light. 3. Matter waves are not real waves but are probability waves only, whereas EM waves are real waves with periodic variation of E and B vector in them. 4. Matter waves are produced by a motion of particle and are independent of charges, whereas EM waves are produced by motion of charged particle only. (g) A superconductor has a critical temperature 3.7°K at zero magnetic field. At O°K the critical magnetic field is 0.0306 Tesla. What is the critical magnetic field at temperature 2.0°K? We have, \begin{array}{l}H_c\left(T\right)=H_c\left(o\right)\left[1-\left(\frac T{T_c}\right)\right]^2\\\;\;\;\;\;\;\;\;\;\;=0.0306\left[1-\left(\frac2{3.7}\right)\right]^2\\H_c\left(T\right)=0.00646\;K\\\end{array} 2. (a) Show that the diameter of Newton's nth dark ring is proportional to square root of ring number. In Newton's rings experiment the diameter of 5th dark ring was 0.336 cm and that of 15th dark ring was 0.590 cm. Calculate the radius of curvature of plano-convex lens if wavelength of light used is 5890 Å. [8M] Let R be the radius of curvature of the lens. Let a dark ring be located at the point B. The thickness of the air film at B is t and AE = r, the radius of the ring at thickness t. By using the Pythagoras theorem to triangle AEO we get, AO2 =AE2+EO2 R2 = r2+ (R-t)2 = r2 + R2 – 2Rt + t2 r2 = 2Rt or t = r2/2R  ,As t is very small, t2 can be neglected Using relation 2t = nλ, for dark rings. r2 = nλR r=\sqrt{n\lambda R} Dark ring. The Diameter of dark ring is given by D_n=\sqrt{4n\lambda R} Since, λ and  R are constants D_n\alpha\sqrt n i.e. the diameter of the Newton's nth dark ring is directly proportional to square root of the ring number. we have,

\begin{array}{l}R=\frac{D_n^2}{4n\lambda}\\\;\;\;=\frac{\left(0.336\times10^{-2}\right)^2}{4\times5\times5.89\times10^{-7}}\\\;\;\;=0.000958\times10^3\\\;\;\;=95.8\times10^{-2}m\\\;\;\;=95.8\;cm\end{array}<a href="b">/latex</a> Derive an expression for numerical aperture of step index optical fiber. What are the advantages of using an optical fiber? [7M] The main function of the optical fiber is to accept and transmit as much as light from the source as possible. The light gathering ability of the fiber depends on two factors, namely core size and numerical aperture. The numerical aperture determined by the acceptance angle and fractional refractive index change. Acceptance angle and acceptance cone image…… Let us consider an optical fiber into which light is launched. Let refractive index of core be µ1 and refractive index of cladding be µ2.The refractive index of cladding is less than refractive index of core. Let µ0 be the refractive index of the medium from which light is launched into the fiber. Let a light ray enter the fiber at an angle θi to the axis of the fiber. The rays refracts at an angle θr and strike the core cladding interface at an angle [latex] \phi

. If \phi is greater than the critical angle \phic, the rays undergoes total internal reflection at the interface, since µ1>µ2 as long as the angle \phi is greater than critical angle \phic, the light will stay within the fiber as shown figure.

Let us now compute the incident angle θi for which  \phi\;\geq\;\phi_c such that light rebounds within the
fiber. Therefore From Snell’s law,

\begin{array}{l}\frac{\sin\theta_i}{\sin\theta_r}=\frac{\mu_1}{\mu_2}\\\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_2}\sin\theta_r-----\left(i\right)\end{array}

If θi is increased beyond a limit, \phi will drop below the critical value \phic and the ray escape from the sides walls of the fiber.

\sin\theta_r=\sin\left(90-\phi\right)=\cos\phi\;-----\left(ii\right)

Using equation 1 and 2

\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi

when \phi=\phi_c

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi_c-----\left(iii\right)

But by the condition of total internal reflection

\sin\phi_c=\frac{\displaystyle\mu_2}{\displaystyle\mu_1}

Now, Squaring both side

\begin{array}{l}\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\;;\;1-\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\\\cos^2\phi_c=\frac{\mu_1^2-\mu_2^2}{\mu_1^2}\;;\;\cos^2\phi_c=\frac{\displaystyle\mu_1^2-\mu_2^2}{\displaystyle\mu_1^2}\\\cos\phi_c=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_1}-----\left(iv\right)\end{array}

Put equation (iv) in equation (3), we get,

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_0}

Incident ray is launched from air medium for which, \mu_0=1 and substituting \theta_{i\left(max\right)}=\theta_0

we get,

\begin{array}{l}\sin\theta_0=\sqrt{\mu_1^2-\mu_2^2}\\\theta_0=\sin^{-1}=\sqrt{\mu_1^2-\mu_2^2}\end{array}
The angle θ0 is called the acceptance angle of the fiber.

Acceptance angle: It is defined as the maximum angle that a light ray can have relative to the axis of the fiber and propagate down the fiber.

Acceptance Cone: The light ray contained within the cone having a full angle 2θ0 are accepted and transmitted along the fiber. Therefore the cone is called the acceptance cone.

Fractional refractive index change
The fractional difference Δ between the refractive indices of the core and the cladding is known as Fractional refractive index change.

\triangle=\frac{\displaystyle\mu_1-\mu_2}{\mu_1}

Numerical aperture: The numerical aperture (NA) is defined as the acceptance angle.

Thus

\begin{array}{l}N.A.=\sin\theta_0\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}{2\mu_1}}2\mu_1\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1-\mu_2\right)}{\mu_1}}2\mu_1^2\;\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1+\mu_2}2=\mu_1\right)\\N.A.=\mu_1\sqrt{2\triangle}---\left(i\right)\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1-\mu_2}{\mu_1}=\triangle\right)\end{array}

Numerical aperture determines the light gathering ability of the fiber. It is measure of the amount of the light that can be accepted by fiber. It is seen from equation (i) that NA. dependent only on the refractive indices of the core and cladding materials. Its value range from 0.13 to 0.50. A large NA implies that a fiber will accept large amount of light from the source.

  1. (a) Explain construction and working of He-Ne laser. What are its merits ? [8M]

Construction: It consists of a glass discharge tube of about 30 cm long and 1 cm diameter. It is filled with mixture of He and Ne gas in the ratio of 10:1. Two electrodes are provided to produce discharge in the gas and connected to high voltage power supply. The tube is hermetically sealed by inclined windows arranged at its two ends. On the axis of the tube two reflectors are fixed which constitute optical resonator cavity. The distance between the mirrors is adjusted such that it supports standing wave pattern.

Image…..

Working:
The energy level diagram is shown in figure. It employs four level pumping scheme. When the power supply is put on; due to electric field the gas is ionized producing electrons and ions of gas. The electrons and ions are accelerated towards anode and cathode. The electrons transfer their energy to He ions due to collision and are excited to higher state which lie at 19.81 eV and 20.16 eV above the ground state. He- atoms are lighter and hence readily excited. These levels E2 and E3 are metastable states and hence the excited He atoms do not return to ground states through spontaneous emission process. However, the He atoms return to ground states by transferring their energy to Ne atoms through collision there by exciting Ne atoms to energy levels E4 (18.70 eV) and E6 (20.66 eV). This becomes possible because two colliding atoms have identical energy states.

Image…….

The energy difference of 0.05eV is provided by K.E. acquired by He ions. This transfer of energy from He atoms to Ne atoms take place because E6 and E4 energy levels have nearly same values as that of He atoms. Ne atoms do not get excited by directly absorbing energy from pump since they are heavier than He ions. Ne atoms are the active centers and the role of He atoms is to excite Ne atoms which causes population inversion. To ensure proper energy transfer from He to Ne atoms; the proportion of He and Ne atoms is chosen as 10 : 1. This also reduces probability of reverse transfer of energy from Ne atoms to He atoms. The excited states of helium atoms are metastable states hence if the path for their decay is not provided they tend to accumulate and subsequently no He ions are available for excitation of Ne atoms. Decay path for He ions is provided by making the tube narrower so that the excited He atoms collide with walls of tube and return to their ground states.

The E6 and E4 energy states of Ne atoms are metastable states, therefore Ne atoms accumulate in these two states and population inversion between is achieved between E6 - E5, E3 and E4 -E3 levels.
Therefore three LASER transition takes place which are:

  1. E6 to E3 transition: giving red color LASER beam of wavelength 6328 Å
  2. E6 to E5 transition: giving LASER beam in IR region with wavelength 33900 Å
  3. E4 to E3 transition: giving LASER beam in IR region with wavelength 11500 Å

From terminal levels E5 and E3, Ne atoms make spontaneous emission and shows downward transition to E2 level. As E2 is meta-stable state and therefore Ne atoms again accumulate at this level. Atoms from E2 state are required to be brought to ground state E1 because if they accommodate in E2 state, number of atoms in ground state will decrease fast it becomes difficult to maintain population inversion continuously.

In order to avoid this, discharge tube is made narrow due to which probability of collision of atoms with walls increases and atoms can give up excess energy and can return back to ground state easily. This LASER is used as a monochromatic source in interferometry, LASER printers, bar code reading etc. They are also used as reference beam in surveying, alignment in laying pipes etc.

(b) Derive the condition for a thin transparent film of constant thickness to appear bright mid dark when viewed in reflected fight. [7M]

When a monochromatic light beam is incident on a transparent parallel thin film of uniform
thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern.
These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially
reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected
from lower surface of the film along MD and finally emerges along MK
These rays BC and DE interfere and interference fringes are produced. The intensity at any Point
depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and
Determination of path difference between the rays BM and MD. As shown in fig. 1

The geometric path difference between ray 1 and ray 2 = MF + FD - BH

Optical path difference = Δ

\begin{array}{l}=\mu(BM+FD)-\mu BH\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array} \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\Further,\end{array} \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}

But MG=t

BM=\frac t{\cos\;r}\;……….\left(2\right)

Similarly

MD=\frac t{\cos\;r}\;……….\left(3\right)

Now, In ΔBHD

\begin{array}{l}\sin\;i=\frac{BH}{BD}\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\BD=BG+GD\;\;………\left(5\right)\end{array}

But
Therefore, In ΔBGM

\begin{array}{l}\tan\;r=\frac{BG}{MG}\\BG=MG\;\tan\;r\\BG=t\;\tan\;r\end{array}

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

\begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\BD=2t\;\tan\;r\end{array}

Therefore, equation 4 becomes,

BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)

Put equation 2,3 and 6 in equation 1 we get

\begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}

But

\begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\sin\;i=\;\mu\;\sin\;r\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\triangle=2\;\mu t\;\cos\;r\end{array}

When light is reflected from the surface of an optically denser medium, a phase change π is
introduced. Correspondingly, path difference λ/2 is introduced, therefore
The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:

  1. When path difference ∆ = n λ, where n = 0,1,2,………
    Constructive interference takes place and film appears bright.
    Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
    2μ t cos r = (2n +1) λ/2
  2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
    Destructive interference takes place and film appears dark.
    Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
    2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore,
2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.

  1. (a) Calculate the maximum order of diffraction maxima seen from a plane diffraction grating having 5500 lines per cm if light of wavelength 5896 Å falls normally on it. [5M]

We have,

\begin{array}{l}\left(a+b\right)\;\sin\theta=n\lambda\\n=\left(a+b\right)\;\sin\theta\lambda\end{array}

For maximum value of n, {\left[\sin\theta\right]}_{max}=1

\begin{array}{l}n=\frac{a+b}\lambda\\\;\;\;=\frac{\displaystyle\frac{1\;cm}{5500}}{5.893\times10^{-5}cm}\\\;\;\;=\frac1{3.24\times10^{-1}}\\\;\;\;=3.08\\3.08\approx3\;orders\end{array}

(b) Derive Schrodinger's time-independent wave equation. [5M]

Schrodinger’s wave equation determines the motion of atomic particle. Consider a microparticle in motion which is associated with wave function Ψ. The Ψ represents the field of the wave, say. The classical wave equation is of the form,

\frac{\partial^2y}{\partial x^2}=\frac1{\nu^2}\frac{\displaystyle\partial^2y}{\displaystyle\partial t^2}

A solution of this equation is, y\left(x,\;t\right)=Ae^{i\left(kx-wt\right)}

For a microparticle the w and k can be replaced with E and p using Einstein and de-Broglie relations as,

E=\hslash\omega\;and\;p=\hslash k,

Also replacing y(x, t) by \psi(x, t) we may write

\psi\left(x,\;t\right)=Ae^\frac{-\left(Et-px\right)}ℏ

Differentiating with respect to t, we get,

\frac{\partial\psi}{\partial t}=-\frac{i}{\hslash}E\psi

Differentiating twice with respect to x, we get,

\frac{\partial^2\psi}{\partial x^2}=-\frac{p^{2}}{\hslash}\psi

For a free particle we have

E=\frac{p^2}{2m}

and in case of a particle moving in force field characterized by potential energy V, we have

\frac{p^2}{2m}=E-V.

Multiplying above equation by \psi,

\frac{p^2}{2m}\psi=E\psi-V\psi

Substituting for E\psi\;and\;p^2\psi, and  rearranging we get

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=i\hslash\frac{\partial \psi }{\partial t}

Which is time dependent Schrodinger's wave equation, where i\hslash\frac\partial{\partial t}=E, the energy operator.

Therefore,

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=E\psi

This is time independent Schrodinger's wave equation.

(c) Define the term superconductivity. Show that in the superconducting state the material is perfectly diamagnetic. [5M]

Superconductivity: The sudden disappearance of electrical resistance in materials below a certain temperature is known as superconductivity. The materials that exhibit superconductivity and which are in the superconducting state are called superconductors. The temperature at which a normal material turns into a superconductor is called as critical temperature, Tc. Every superconductor has it’s own critical temperature at which it passes over into superconducting state. The superconducting transition is sharp for chemically pure and structurally pure specimen but broad for impure specimens and with structural defects. In 1933 Meissener and Oschsenfield discovered that a superconductor completely expels the magnetic field lines that were initially penetrating it in it’s normal state. This property is independent of the path by which the superconducting state is reached, as shown in the figure. Suppose that magnetic field is applied first to the sample in normal state. The sample is them cooled below Tc, in the presence of magnetic field. It is expected that the magnetic flux through the sample remains unchanged. But, Meissener and Oschsenfield found that magnetic flux was totally expelled by the sample as it becomes superconducting. The expulsion of the magnetic field during the transition from the normal to the superconducting state is called as Meissener effect.

The effect is reversible. When temperature is raised from below Tc, the flux suddenly starts penetrating the specimen at T=Tc as a result of which the material returns back to the normal state. The magnetic induction inside the specimen is given by, B=\mu_0\left(H+M\right)

Where, H is the external magnetic field and M is the magnetization produced in the specimen. At, T<T_c,\;B=0\;and\;\mu_0\left(H+M\right)=0 superconducting state.

Therefore, M = -H

The susceptibility of the material is. x = M/H = -1

Thus, the superconducting state is characterized by perfect diamagnetism. Meissener effect shows that in superconductor not only dB/dt =0 but also B=0. The superconducting state is a characteristic thermodynamic phase of a substance in which material cannot sustain steady electric and magnetic field. Meissener effect is the standard test which conclusively proves that the material is in superconducting state or not. Because of diamagnetism superconducting material strongly repel the external magnets. It leads to levitation effect in which a magnet hovers over superconducting material and also suspension effect in which a chip of superconducting material hangs beneath a magnet.

  1. (a) A slit of width 0.3 mm is illuminated by a light of wavelength 5890 Å. A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern. Calculate the distance between first dark and the next bright fringe form the axis. [5M]

The distance of first dark band from the centre is λD/a and next bright band is 3λD/2a.
Thus, the distance between first dark and next bright will be,

\begin{array}{l}x=\frac{\lambda D}{2a}\\\;\;\;=\frac{5.89\times10^{-5}cm\times40cm}{2\times0.03cm}\\\;\;\;=0.0392\;cm\end{array}

(Note: the focal length f is nearly equal to D, when lens is placed close to the slits.)

(b) An electron is accelerated through 1000 volts and is reflected from a crystal. The first order reflection occurs when glancing angle is 70°. Calculate the interplanar spacing of a crystal. [5M]

We have,

2d\;\sin\theta=n\lambda

Where \begin{array}{l}\lambda=\frac{12.26}{\sqrt v}\overset\circ A\\\;\;\;=\frac{12.26}{\sqrt{1000}}\\\;\;\;=0.3876\;\overset\circ A\end{array}

Thus, \begin{array}{l}d=\frac{n\lambda}{2\;\sin\theta}\\\;\;\;=\frac{1\times0.3876\;\overset\circ A}{2\;\sin\;70}\\\;\;\;=0.2062\;\overset\circ A\end{array}

(c) Explain construction and working of Atomic Force Microscope. [5M]

Atomic force microscopy (AFM) or scanning force microscopy (SFM) is a very high-resolution
type of scanning probe microscopy, with demonstrated resolution on the order of fractions of a nanometer, more than 1000 times better than the optical diffraction limit. The AFM is one of the foremost  tools for imaging, measuring, and manipulating matter at the nanoscale.

The information is gathered by "feeling" the surface with a mechanical probe. Piezoelectric
elements that facilitate tiny but accurate and precise movements on (electronic) command enable the very precise scanning. In some variations, electric potentials can also be scanned using conducting cantilevers. In newer more advanced versions, currents can even be passed through the tip to probe the electrical conductivity or transport of the underlying surface, but this is much more challenging with very few research groups reporting reliable data.

The AFM consists of a cantilever with a sharp tip (probe) at its end that is used to scan the specimen surface. The cantilever is typically silicon or silicon nitride with a tip radius of curvature on the order of nanometers. When the tip is brought into proximity of a sample surface, forces between the tip and the sample lead to a deflection of the cantilever according to Hooke's law.

Depending on the situation, forces that are measured in AFM include mechanical contact force, van der Waals forces, capillary forces, chemical bonding, electrostatic forces, magnetic forces (see magnetic force microscope, MFM), Casimir forces, solvation forces, etc. Along with force, additional quantities may simultaneously be measured through the use of specialized types of probe.

If the tip was scanned at a constant height, a risk would exist that the tip collides with the surface, causing damage. Hence, in most cases a feedback mechanism is employed to adjust the tip-to-sample distance to maintain a constant force between the tip and the sample. Traditionally, the sample is mounted on a piezoelectric tube that can move the sample in the z direction for maintaining a constant force, and the x and y directions for scanning the sample.

Alternatively a 'tripod' configuration of three piezo crystals may be employed, with each
responsible for scanning in the x,y and z directions. This eliminates some of the distortion effects seen with a tube scanner. In newer designs, the tip is mounted on a vertical piezo scanner while the sample is being scanned in X and Y using another piezo block. The resulting map of the area z = f(x, y) represents the topography of the sample.

The AFM can be operated in a number of modes, depending on the application. In general,
possible imaging modes are divided into static (also called contact) modes and a variety of dynamic (or non-contact) modes where the cantilever is vibrated.

Image……

  1. (a) State Heisenber's uncertainty principle. Show that electron cannot pre-exist in free state in a nucleus. [5M]

Statement: “It is impossible to determine exact position and exact momentum both of a particle simultaneously with unlimited accuracy.”

It means that, if the position of the particle is determined accurately then there is uncertainty in the determination of momentum and if the momentum of the particle is determined accurately then there is uncertainty in the determination of its position.

Mathematically, “The product of uncertainties in simultaneous measurement of position and momentum of a microscopic particle is always of the order of Planck’s constant or greater than or equal to h/2π”.
Thus mathematically,

Δx Δpx ≥ ђ [ ђ = h/2 π ]

Or

Δx Δpx ≥ h/2π

This is called as Heisenberg’s uncertainty relation. Where, Δx = Uncertainty in measurement of position, ΔpX = uncertainty in measurement of momentum.
If Δx is small i.e. position is determined more accurately then ΔpX will be large i.e. momentum is determined less accurately and vice-versa.
If electron is inside the nucleus, we will get it within its diameter. i.e. \triangle x=2\times10^{-14}.
Now, Δx Δp = h/2π or Δp ≥ h/2π Δx ≈ 5 x 10-21 kgm/s.

This, minimum energy,

\begin{array}{l}E=\;p.c\;=\;mc.c\;\;\\\;\;\;=\;3.2\;x\;10-12\;J\;\\\;\;\;=\;20\;MeV.\end{array}

Thus if electron is inside nucleus its energy must be about 20 MeV. But, the maximum energy of electron found is 3 MeV, hence electron cannot reside the nucleus.

(b) Draw a labelled diagram and explain construction and working of CRT. [5M]

Cathode Ray Tube is an important part of CRO and is said to be the heart of CRO. It generates the electron beam, focuses it and accelerates it towards the fluorescent screen. It mainly consists of (i)Electron gun assembly, (ii) Deflection plate assembly, and (iii) Fluorescent screen as shown in figure.

1 Electron gun assembly
The function of electron gun is to produce a sharply focussed beam of electrons, which are accelerated to a high velocity. It consists of a heater, cathode, control grid, pre accelerating anode, focusing anode and an accelerating anode.

The cathode is in the form of a cylinder. The electrons emitted from this indirectly heated cathode come out through a small hole and enter control grid. The control grid is negatively biased cylinder with a centrally located hole. It controls the number of electrons reaching the anode and thus the intensity of electron beam is controlled by the grid. These electrons are accelerated by a high positive potential to preaccelerating and accelerating anodes

Image…….

2 Deflection plate assembly:
It consists of two pairs of deflecting plates for deflecting the beam of electron both in the vertical and horizontal directions. One pair of plates, called the X-plates, is mounted vertically and it deflects the electron beam in the horizontal direction due to the electric field produced in horizontal plane by applying certain potential across these plates. Another pair of plates called the Y-plates is mounted horizontally which produces the electric field in a vertical plane, after applying a potential difference across them. This produces vertical deflection of the electron beam. The plates are flared so as to allow the beam to pass
through them without striking the plates.

3 Fluorescent Screen
The screen of CRT is coated with a fluorescent material like Zink orthosilicate, in general called as phosphor. It absorbs the kinetic energy of electrons striking it and re-emits in the form of light. Hence the screen is called as fluorescent screen. The colour of the trace on the screen depends on the type of material.

All these parts of CRT are enclosed in a funnel shaped highly evacuated glass envelope. The
inner surface of the flared part of this envelope is coated with an aqueous solution of graphite called aquadag, which is internally connected to the accelerating anode. This layer performs two functions-
(1) It accelerates the electron beam, (2) It collects the secondary electrons produced by secondary emission when electron beam strikes the screen.

(c) Exploits top down and bottom up approaches to prepare nanomaterials. [5M]

Engineered nano-objects and nanomaterials intended for industrial use may be synthesized using two different approaches known as the op down Approach and the Bottom Up Approach.

  1. Top-down approach
    The most common top-down approach to fabrication involves lithographic patterning techniques using short-wavelength optical sources. A key advantage of the top-down approach as developed in the fabrication of integrated circuits is that the parts are both patterned and built in place, so that no assembly step is needed. Optical lithography is a relatively mature field because of the high degree of refinement in microelectronic chip manufacturing, with current short-wavelength optical lithography techniques reaching dimensions just below 100 nanometers (the traditional threshold definition of the nanoscale).

Shorter-wavelength sources, such as extreme ultraviolet and X-ray, are being developed to allow lithographic printing techniques to reach dimensions from 10 to 100 nanometers. Scanning beam techniques such as electron-beam lithography provide patterns down to about 20 nanometers. Here the pattern is written by sweeping a finely focused electron beam across the surface. Focused ion beams are also used for direct processing and patterning of wafers, although with somewhat less resolution than in electron-beam lithography. Still-smaller features are obtained by using scanning probes to deposit or remove thin layers. Mechanical printing techniques—nanoscale imprinting, stamping, and molding—have been extended to the surprisingly small dimensions of about 20 to 40 nanometers.

  1. Bottom-up approach
    Bottom-up or self-assembly approach to nanofabrication use chemical or physical forces operating at the nanoscale to assemble basic units into larger structures. As component size decreases in nanofabrication, bottom-up approaches provide an increasingly important complement to top-down techniques. Inspiration for bottom-up approach comes from biological systems, where nature has harnessed chemical forces to create essentially all the structures needed by life. Researchers hope to replicate nature‘s ability to produce small clusters of specific atoms, which can then self-assemble into more-elaborate structures. A number of bottom-up approaches have been developed for producing nanoparticles, ranging from condensation of atomic vapors on surfaces to coalescence of atoms in liquids. For example, liquid-phase techniques have been developed to produce size-selected nanoparticles of semiconductor, magnetic, and other materials.

An example of self-assembly that achieves a limited degree of control over both formation and organization is the growth of quantum dots. Indium gallium arsenide (InGaAs) dots can be formed by growing thin layers of InGaAs on GaAs in such a manner that repulsive forces caused by compressive strain in the InGaAs layer results in the formation of isolated quantum dots. After the growth of multiple layer pairs, a fairly uniform spacing of the dots can be achieved. Another example of self-assembly of an intricate structure is the formation of carbon nanotubes under the right set of chemical and temperature conditions.

Image…….B.E. Second Semester All Branches (C.B.S.)
Time: 2 hours
Maximum marks: 60

Notes :
1.Question No. 1 is compulsory.
2. Attempt any three questions from Q.2 to Q.6.

  1. Used suitable data wherever required.
  2. Figures to the right indicate full mark.
  3. Attempt any five of the following :- [15M]
    (a) Comment on colours in a soap film in sunlight.

If sunlight is incident on a soap film, the optical path difference will vary from one color to other due to change of wavelength with color. Therefore the film will appear to be colored and the color seen will be of that rays for which constructive interference occurs. If the sunlight is not parallel then the optical path difference will change due to change in angle of incidence. Hence, the film will show different colors when viewed from different direction.

(b) What is Rayleigh's criterion of resolution ? Define resolving power of a grating.

When an object is seen through a optical instrument, diffraction occurs as the light passes through the aperture. Hence in actual practice the diffraction pattern of the object is seen when viewed through any optical instrument. Because of which there is a limit on the ability of that instrument to view the two closely spaced point objects separately.

According to Rayleigh’s criterion the closely spaced point objects are said to be just resolved, or can be seen separately, when the principle maxima of one objects fall upon the first minimum of second object and vice versa. If the distance decreases then they are not resolved and if distance increased then they are said to be well resolved.

Resolving power of grating can be defined as the reciprocal of angular or linear distance for which two spectral lines can be seen separately and is given by

R=λ/Δλ.

(c) Calculate V number for an optical fiber having numerical aperture 0.25 and core diameter 20 μm if it is operated at 1.55 μm.

Given,
NA=0.25, d= 20 x 10-6 m, λo= 1.55 x 10-6 m,

We have,

\begin{array}{l}V=\frac{\pi d}\lambda\left[NA\right]\\\;\;\;=\frac{3.14\times20\times10^{-6}}{1.55\times10^{-6}}\times0.25\\\;\;\;=10.129\end{array}

(d) Compare light from ordinary source with laser light.

1) Light from ordinary source is not directional, whereas laser light is highly directional.

2) Ordinary light is poly-chromatic whereas LASER is highly monochromatic and contains a very
narrow range of the order of a few Angstroms (<10 Å ). 3) The intensity of ordinary light is very small whereas LASER emits light in the form of a narrow beam with its energy concentrated in a small region of space .Therefore, the beam intensity would be tremendously large and stays constant with distance since it travels in the form of plane waves. 4) Ordinary light is highly divergent whereas Light from a LASER propagates in the form of plane waves and it has very small divergence. 5) ordinary light is incoherent light since they emit light waves of random wavelengths having random phases. While, the waves emitted by LASER source are in phase and having same frequency. Therefore, light emitted by LASER is highly coherent. (e) How phase difference between two signals is measured using CRO? When two sine waves oscillating in mutually perpendicular direction are of same frequency, the Lissajous pattern takes the form of an ellipse as shown in the figure. This ellipse is used to determine the phase difference between two signals, by using the formula \phi=\sin^{-1}\left(\frac{AA'}{BB'}\right) By measuring the length AA’ and BB’ on the oscillogram and substituting the values in the above formula phase difference can be determined. Images……. (f) What are the properties of matter waves? 1. Wavelength of Matter waves depends upon mass and velocity of particle \left(\lambda=\frac h{m\nu}\right), where as wavelength of EM waves depends upon energy \left(\lambda=\frac{hc}E\right). 2. Velocity of matter wave depend upon velocity of particle and greater than velocity of light, whereas velocity of EM wave is constant equal to velocity of light. 3. Matter waves are not real waves but are probability waves only, whereas EM waves are real waves with periodic variation of E and B vector in them. 4. Matter waves are produced by a motion of particle and are independent of charges, whereas EM waves are produced by motion of charged particle only. (g) A superconductor has a critical temperature 3.7°K at zero magnetic field. At O°K the critical magnetic field is 0.0306 Tesla. What is the critical magnetic field at temperature 2.0°K? We have, \begin{array}{l}H_c\left(T\right)=H_c\left(o\right)\left[1-\left(\frac T{T_c}\right)\right]^2\\\;\;\;\;\;\;\;\;\;\;=0.0306\left[1-\left(\frac2{3.7}\right)\right]^2\\H_c\left(T\right)=0.00646\;K\\\end{array} 2. (a) Show that the diameter of Newton's nth dark ring is proportional to square root of ring number. In Newton's rings experiment the diameter of 5th dark ring was 0.336 cm and that of 15th dark ring was 0.590 cm. Calculate the radius of curvature of plano-convex lens if wavelength of light used is 5890 Å. [8M] Let R be the radius of curvature of the lens. Let a dark ring be located at the point B. The thickness of the air film at B is t and AE = r, the radius of the ring at thickness t. By using the Pythagoras theorem to triangle AEO we get, AO2 =AE2+EO2 R2 = r2+ (R-t)2 = r2 + R2 – 2Rt + t2 r2 = 2Rt or t = r2/2R  ,As t is very small, t2 can be neglected Using relation 2t = nλ, for dark rings. r2 = nλR r=\sqrt{n\lambda R} Dark ring. The Diameter of dark ring is given by D_n=\sqrt{4n\lambda R} Since, λ and  R are constants D_n\alpha\sqrt n i.e. the diameter of the Newton's nth dark ring is directly proportional to square root of the ring number. we have,

\begin{array}{l}R=\frac{D_n^2}{4n\lambda}\\\;\;\;=\frac{\left(0.336\times10^{-2}\right)^2}{4\times5\times5.89\times10^{-7}}\\\;\;\;=0.000958\times10^3\\\;\;\;=95.8\times10^{-2}m\\\;\;\;=95.8\;cm\end{array}<a href="b">/latex</a> Derive an expression for numerical aperture of step index optical fiber. What are the advantages of using an optical fiber? [7M] The main function of the optical fiber is to accept and transmit as much as light from the source as possible. The light gathering ability of the fiber depends on two factors, namely core size and numerical aperture. The numerical aperture determined by the acceptance angle and fractional refractive index change. Acceptance angle and acceptance cone image…… Let us consider an optical fiber into which light is launched. Let refractive index of core be µ1 and refractive index of cladding be µ2.The refractive index of cladding is less than refractive index of core. Let µ0 be the refractive index of the medium from which light is launched into the fiber. Let a light ray enter the fiber at an angle θi to the axis of the fiber. The rays refracts at an angle θr and strike the core cladding interface at an angle [latex] \phi

. If \phi is greater than the critical angle \phic, the rays undergoes total internal reflection at the interface, since µ1>µ2 as long as the angle \phi is greater than critical angle \phic, the light will stay within the fiber as shown figure.

Let us now compute the incident angle θi for which  \phi\;\geq\;\phi_c such that light rebounds within the
fiber. Therefore From Snell’s law,

\begin{array}{l}\frac{\sin\theta_i}{\sin\theta_r}=\frac{\mu_1}{\mu_2}\\\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_2}\sin\theta_r-----\left(i\right)\end{array}

If θi is increased beyond a limit, \phi will drop below the critical value \phic and the ray escape from the sides walls of the fiber.

\sin\theta_r=\sin\left(90-\phi\right)=\cos\phi\;-----\left(ii\right)

Using equation 1 and 2

\sin\theta_i=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi

when \phi=\phi_c

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\mu_1}{\displaystyle\mu_0}\cos\phi_c-----\left(iii\right)

But by the condition of total internal reflection

\sin\phi_c=\frac{\displaystyle\mu_2}{\displaystyle\mu_1}

Now, Squaring both side

\begin{array}{l}\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\;;\;1-\sin^2\phi_c=\frac{\displaystyle\mu_2^2}{\displaystyle\mu_1^2}\\\cos^2\phi_c=\frac{\mu_1^2-\mu_2^2}{\mu_1^2}\;;\;\cos^2\phi_c=\frac{\displaystyle\mu_1^2-\mu_2^2}{\displaystyle\mu_1^2}\\\cos\phi_c=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_1}-----\left(iv\right)\end{array}

Put equation (iv) in equation (3), we get,

{\left[\sin\theta_i\right]}_{\left(max\right)}=\frac{\displaystyle\sqrt{\mu_1^2-\mu_2^2}}{\displaystyle\mu_0}

Incident ray is launched from air medium for which, \mu_0=1 and substituting \theta_{i\left(max\right)}=\theta_0

we get,

\begin{array}{l}\sin\theta_0=\sqrt{\mu_1^2-\mu_2^2}\\\theta_0=\sin^{-1}=\sqrt{\mu_1^2-\mu_2^2}\end{array}
The angle θ0 is called the acceptance angle of the fiber.

Acceptance angle: It is defined as the maximum angle that a light ray can have relative to the axis of the fiber and propagate down the fiber.

Acceptance Cone: The light ray contained within the cone having a full angle 2θ0 are accepted and transmitted along the fiber. Therefore the cone is called the acceptance cone.

Fractional refractive index change
The fractional difference Δ between the refractive indices of the core and the cladding is known as Fractional refractive index change.

\triangle=\frac{\displaystyle\mu_1-\mu_2}{\mu_1}

Numerical aperture: The numerical aperture (NA) is defined as the acceptance angle.

Thus

\begin{array}{l}N.A.=\sin\theta_0\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\mu_1^2-\mu_2^2}\\\;\;\;\;\;\;\;\;\;=\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}{2\mu_1}}2\mu_1\\\;\;\;\;\;\;\;\;\;=\sqrt{\frac{\displaystyle\left(\mu_1-\mu_2\right)}{\mu_1}}2\mu_1^2\;\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1+\mu_2}2=\mu_1\right)\\N.A.=\mu_1\sqrt{2\triangle}---\left(i\right)\;\;\;As\;\;\;\;\left(\frac{\displaystyle\mu_1-\mu_2}{\mu_1}=\triangle\right)\end{array}

Numerical aperture determines the light gathering ability of the fiber. It is measure of the amount of the light that can be accepted by fiber. It is seen from equation (i) that NA. dependent only on the refractive indices of the core and cladding materials. Its value range from 0.13 to 0.50. A large NA implies that a fiber will accept large amount of light from the source.

  1. (a) Explain construction and working of He-Ne laser. What are its merits ? [8M]

Construction: It consists of a glass discharge tube of about 30 cm long and 1 cm diameter. It is filled with mixture of He and Ne gas in the ratio of 10:1. Two electrodes are provided to produce discharge in the gas and connected to high voltage power supply. The tube is hermetically sealed by inclined windows arranged at its two ends. On the axis of the tube two reflectors are fixed which constitute optical resonator cavity. The distance between the mirrors is adjusted such that it supports standing wave pattern.

Image…..

Working:
The energy level diagram is shown in figure. It employs four level pumping scheme. When the power supply is put on; due to electric field the gas is ionized producing electrons and ions of gas. The electrons and ions are accelerated towards anode and cathode. The electrons transfer their energy to He ions due to collision and are excited to higher state which lie at 19.81 eV and 20.16 eV above the ground state. He- atoms are lighter and hence readily excited. These levels E2 and E3 are metastable states and hence the excited He atoms do not return to ground states through spontaneous emission process. However, the He atoms return to ground states by transferring their energy to Ne atoms through collision there by exciting Ne atoms to energy levels E4 (18.70 eV) and E6 (20.66 eV). This becomes possible because two colliding atoms have identical energy states.

Image…….

The energy difference of 0.05eV is provided by K.E. acquired by He ions. This transfer of energy from He atoms to Ne atoms take place because E6 and E4 energy levels have nearly same values as that of He atoms. Ne atoms do not get excited by directly absorbing energy from pump since they are heavier than He ions. Ne atoms are the active centers and the role of He atoms is to excite Ne atoms which causes population inversion. To ensure proper energy transfer from He to Ne atoms; the proportion of He and Ne atoms is chosen as 10 : 1. This also reduces probability of reverse transfer of energy from Ne atoms to He atoms. The excited states of helium atoms are metastable states hence if the path for their decay is not provided they tend to accumulate and subsequently no He ions are available for excitation of Ne atoms. Decay path for He ions is provided by making the tube narrower so that the excited He atoms collide with walls of tube and return to their ground states.

The E6 and E4 energy states of Ne atoms are metastable states, therefore Ne atoms accumulate in these two states and population inversion between is achieved between E6 - E5, E3 and E4 -E3 levels.
Therefore three LASER transition takes place which are:

  1. E6 to E3 transition: giving red color LASER beam of wavelength 6328 Å
  2. E6 to E5 transition: giving LASER beam in IR region with wavelength 33900 Å
  3. E4 to E3 transition: giving LASER beam in IR region with wavelength 11500 Å

From terminal levels E5 and E3, Ne atoms make spontaneous emission and shows downward transition to E2 level. As E2 is meta-stable state and therefore Ne atoms again accumulate at this level. Atoms from E2 state are required to be brought to ground state E1 because if they accommodate in E2 state, number of atoms in ground state will decrease fast it becomes difficult to maintain population inversion continuously.

In order to avoid this, discharge tube is made narrow due to which probability of collision of atoms with walls increases and atoms can give up excess energy and can return back to ground state easily. This LASER is used as a monochromatic source in interferometry, LASER printers, bar code reading etc. They are also used as reference beam in surveying, alignment in laying pipes etc.

(b) Derive the condition for a thin transparent film of constant thickness to appear bright mid dark when viewed in reflected fight. [7M]

When a monochromatic light beam is incident on a transparent parallel thin film of uniform
thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern.
These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially
reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected
from lower surface of the film along MD and finally emerges along MK
These rays BC and DE interfere and interference fringes are produced. The intensity at any Point
depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and
Determination of path difference between the rays BM and MD. As shown in fig. 1

The geometric path difference between ray 1 and ray 2 = MF + FD - BH

Optical path difference = Δ

\begin{array}{l}=\mu(BM+FD)-\mu BH\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array} \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\Further,\end{array} \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}

But MG=t

BM=\frac t{\cos\;r}\;……….\left(2\right)

Similarly

MD=\frac t{\cos\;r}\;……….\left(3\right)

Now, In ΔBHD

\begin{array}{l}\sin\;i=\frac{BH}{BD}\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\BD=BG+GD\;\;………\left(5\right)\end{array}

But
Therefore, In ΔBGM

\begin{array}{l}\tan\;r=\frac{BG}{MG}\\BG=MG\;\tan\;r\\BG=t\;\tan\;r\end{array}

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

\begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\BD=2t\;\tan\;r\end{array}

Therefore, equation 4 becomes,

BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)

Put equation 2,3 and 6 in equation 1 we get

\begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}

But

\begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\sin\;i=\;\mu\;\sin\;r\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\triangle=2\;\mu t\;\cos\;r\end{array}

When light is reflected from the surface of an optically denser medium, a phase change π is
introduced. Correspondingly, path difference λ/2 is introduced, therefore
The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:

  1. When path difference ∆ = n λ, where n = 0,1,2,………
    Constructive interference takes place and film appears bright.
    Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
    2μ t cos r = (2n +1) λ/2
  2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
    Destructive interference takes place and film appears dark.
    Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
    2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore,
2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.

  1. (a) Calculate the maximum order of diffraction maxima seen from a plane diffraction grating having 5500 lines per cm if light of wavelength 5896 Å falls normally on it. [5M]

We have,

\begin{array}{l}\left(a+b\right)\;\sin\theta=n\lambda\\n=\left(a+b\right)\;\sin\theta\lambda\end{array}

For maximum value of n, {\left[\sin\theta\right]}_{max}=1

\begin{array}{l}n=\frac{a+b}\lambda\\\;\;\;=\frac{\displaystyle\frac{1\;cm}{5500}}{5.893\times10^{-5}cm}\\\;\;\;=\frac1{3.24\times10^{-1}}\\\;\;\;=3.08\\3.08\approx3\;orders\end{array}

(b) Derive Schrodinger's time-independent wave equation. [5M]

Schrodinger’s wave equation determines the motion of atomic particle. Consider a microparticle in motion which is associated with wave function Ψ. The Ψ represents the field of the wave, say. The classical wave equation is of the form,

\frac{\partial^2y}{\partial x^2}=\frac1{\nu^2}\frac{\displaystyle\partial^2y}{\displaystyle\partial t^2}

A solution of this equation is, y\left(x,\;t\right)=Ae^{i\left(kx-wt\right)}

For a microparticle the w and k can be replaced with E and p using Einstein and de-Broglie relations as,

E=\hslash\omega\;and\;p=\hslash k,

Also replacing y(x, t) by \psi(x, t) we may write

\psi\left(x,\;t\right)=Ae^\frac{-\left(Et-px\right)}ℏ

Differentiating with respect to t, we get,

\frac{\partial\psi}{\partial t}=-\frac{i}{\hslash}E\psi

Differentiating twice with respect to x, we get,

\frac{\partial^2\psi}{\partial x^2}=-\frac{p^{2}}{\hslash}\psi

For a free particle we have

E=\frac{p^2}{2m}

and in case of a particle moving in force field characterized by potential energy V, we have

\frac{p^2}{2m}=E-V.

Multiplying above equation by \psi,

\frac{p^2}{2m}\psi=E\psi-V\psi

Substituting for E\psi\;and\;p^2\psi, and  rearranging we get

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=i\hslash\frac{\partial \psi }{\partial t}

Which is time dependent Schrodinger's wave equation, where i\hslash\frac\partial{\partial t}=E, the energy operator.

Therefore,

-\frac{\hslash^2\partial^2\psi}{2m\partial x^2}+V\psi=E\psi

This is time independent Schrodinger's wave equation.

(c) Define the term superconductivity. Show that in the superconducting state the material is perfectly diamagnetic. [5M]

Superconductivity: The sudden disappearance of electrical resistance in materials below a certain temperature is known as superconductivity. The materials that exhibit superconductivity and which are in the superconducting state are called superconductors. The temperature at which a normal material turns into a superconductor is called as critical temperature, Tc. Every superconductor has it’s own critical temperature at which it passes over into superconducting state. The superconducting transition is sharp for chemically pure and structurally pure specimen but broad for impure specimens and with structural defects. In 1933 Meissener and Oschsenfield discovered that a superconductor completely expels the magnetic field lines that were initially penetrating it in it’s normal state. This property is independent of the path by which the superconducting state is reached, as shown in the figure. Suppose that magnetic field is applied first to the sample in normal state. The sample is them cooled below Tc, in the presence of magnetic field. It is expected that the magnetic flux through the sample remains unchanged. But, Meissener and Oschsenfield found that magnetic flux was totally expelled by the sample as it becomes superconducting. The expulsion of the magnetic field during the transition from the normal to the superconducting state is called as Meissener effect.

The effect is reversible. When temperature is raised from below Tc, the flux suddenly starts penetrating the specimen at T=Tc as a result of which the material returns back to the normal state. The magnetic induction inside the specimen is given by, B=\mu_0\left(H+M\right)

Where, H is the external magnetic field and M is the magnetization produced in the specimen. At, T<T_c,\;B=0\;and\;\mu_0\left(H+M\right)=0 superconducting state.

Therefore, M = -H

The susceptibility of the material is. x = M/H = -1

Thus, the superconducting state is characterized by perfect diamagnetism. Meissener effect shows that in superconductor not only dB/dt =0 but also B=0. The superconducting state is a characteristic thermodynamic phase of a substance in which material cannot sustain steady electric and magnetic field. Meissener effect is the standard test which conclusively proves that the material is in superconducting state or not. Because of diamagnetism superconducting material strongly repel the external magnets. It leads to levitation effect in which a magnet hovers over superconducting material and also suspension effect in which a chip of superconducting material hangs beneath a magnet.

  1. (a) A slit of width 0.3 mm is illuminated by a light of wavelength 5890 Å. A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern. Calculate the distance between first dark and the next bright fringe form the axis. [5M]

The distance of first dark band from the centre is λD/a and next bright band is 3λD/2a.
Thus, the distance between first dark and next bright will be,

\begin{array}{l}x=\frac{\lambda D}{2a}\\\;\;\;=\frac{5.89\times10^{-5}cm\times40cm}{2\times0.03cm}\\\;\;\;=0.0392\;cm\end{array}

(Note: the focal length f is nearly equal to D, when lens is placed close to the slits.)

(b) An electron is accelerated through 1000 volts and is reflected from a crystal. The first order reflection occurs when glancing angle is 70°. Calculate the interplanar spacing of a crystal. [5M]

We have,

2d\;\sin\theta=n\lambda

Where \begin{array}{l}\lambda=\frac{12.26}{\sqrt v}\overset\circ A\\\;\;\;=\frac{12.26}{\sqrt{1000}}\\\;\;\;=0.3876\;\overset\circ A\end{array}

Thus, \begin{array}{l}d=\frac{n\lambda}{2\;\sin\theta}\\\;\;\;=\frac{1\times0.3876\;\overset\circ A}{2\;\sin\;70}\\\;\;\;=0.2062\;\overset\circ A\end{array}

(c) Explain construction and working of Atomic Force Microscope. [5M]

Atomic force microscopy (AFM) or scanning force microscopy (SFM) is a very high-resolution
type of scanning probe microscopy, with demonstrated resolution on the order of fractions of a nanometer, more than 1000 times better than the optical diffraction limit. The AFM is one of the foremost  tools for imaging, measuring, and manipulating matter at the nanoscale.

The information is gathered by "feeling" the surface with a mechanical probe. Piezoelectric
elements that facilitate tiny but accurate and precise movements on (electronic) command enable the very precise scanning. In some variations, electric potentials can also be scanned using conducting cantilevers. In newer more advanced versions, currents can even be passed through the tip to probe the electrical conductivity or transport of the underlying surface, but this is much more challenging with very few research groups reporting reliable data.

The AFM consists of a cantilever with a sharp tip (probe) at its end that is used to scan the specimen surface. The cantilever is typically silicon or silicon nitride with a tip radius of curvature on the order of nanometers. When the tip is brought into proximity of a sample surface, forces between the tip and the sample lead to a deflection of the cantilever according to Hooke's law.

Depending on the situation, forces that are measured in AFM include mechanical contact force, van der Waals forces, capillary forces, chemical bonding, electrostatic forces, magnetic forces (see magnetic force microscope, MFM), Casimir forces, solvation forces, etc. Along with force, additional quantities may simultaneously be measured through the use of specialized types of probe.

If the tip was scanned at a constant height, a risk would exist that the tip collides with the surface, causing damage. Hence, in most cases a feedback mechanism is employed to adjust the tip-to-sample distance to maintain a constant force between the tip and the sample. Traditionally, the sample is mounted on a piezoelectric tube that can move the sample in the z direction for maintaining a constant force, and the x and y directions for scanning the sample.

Alternatively a 'tripod' configuration of three piezo crystals may be employed, with each
responsible for scanning in the x,y and z directions. This eliminates some of the distortion effects seen with a tube scanner. In newer designs, the tip is mounted on a vertical piezo scanner while the sample is being scanned in X and Y using another piezo block. The resulting map of the area z = f(x, y) represents the topography of the sample.

The AFM can be operated in a number of modes, depending on the application. In general,
possible imaging modes are divided into static (also called contact) modes and a variety of dynamic (or non-contact) modes where the cantilever is vibrated.

Image……

  1. (a) State Heisenber's uncertainty principle. Show that electron cannot pre-exist in free state in a nucleus. [5M]

Statement: “It is impossible to determine exact position and exact momentum both of a particle simultaneously with unlimited accuracy.”

It means that, if the position of the particle is determined accurately then there is uncertainty in the determination of momentum and if the momentum of the particle is determined accurately then there is uncertainty in the determination of its position.

Mathematically, “The product of uncertainties in simultaneous measurement of position and momentum of a microscopic particle is always of the order of Planck’s constant or greater than or equal to h/2π”.
Thus mathematically,

Δx Δpx ≥ ђ [ ђ = h/2 π ]

Or

Δx Δpx ≥ h/2π

This is called as Heisenberg’s uncertainty relation. Where, Δx = Uncertainty in measurement of position, ΔpX = uncertainty in measurement of momentum.
If Δx is small i.e. position is determined more accurately then ΔpX will be large i.e. momentum is determined less accurately and vice-versa.
If electron is inside the nucleus, we will get it within its diameter. i.e. \triangle x=2\times10^{-14}.
Now, Δx Δp = h/2π or Δp ≥ h/2π Δx ≈ 5 x 10-21 kgm/s.

This, minimum energy,

\begin{array}{l}E=\;p.c\;=\;mc.c\;\;\\\;\;\;=\;3.2\;x\;10-12\;J\;\\\;\;\;=\;20\;MeV.\end{array}

Thus if electron is inside nucleus its energy must be about 20 MeV. But, the maximum energy of electron found is 3 MeV, hence electron cannot reside the nucleus.

(b) Draw a labelled diagram and explain construction and working of CRT. [5M]

Cathode Ray Tube is an important part of CRO and is said to be the heart of CRO. It generates the electron beam, focuses it and accelerates it towards the fluorescent screen. It mainly consists of (i)Electron gun assembly, (ii) Deflection plate assembly, and (iii) Fluorescent screen as shown in figure.

1 Electron gun assembly
The function of electron gun is to produce a sharply focussed beam of electrons, which are accelerated to a high velocity. It consists of a heater, cathode, control grid, pre accelerating anode, focusing anode and an accelerating anode.

The cathode is in the form of a cylinder. The electrons emitted from this indirectly heated cathode come out through a small hole and enter control grid. The control grid is negatively biased cylinder with a centrally located hole. It controls the number of electrons reaching the anode and thus the intensity of electron beam is controlled by the grid. These electrons are accelerated by a high positive potential to preaccelerating and accelerating anodes

Image…….

2 Deflection plate assembly:
It consists of two pairs of deflecting plates for deflecting the beam of electron both in the vertical and horizontal directions. One pair of plates, called the X-plates, is mounted vertically and it deflects the electron beam in the horizontal direction due to the electric field produced in horizontal plane by applying certain potential across these plates. Another pair of plates called the Y-plates is mounted horizontally which produces the electric field in a vertical plane, after applying a potential difference across them. This produces vertical deflection of the electron beam. The plates are flared so as to allow the beam to pass
through them without striking the plates.

3 Fluorescent Screen
The screen of CRT is coated with a fluorescent material like Zink orthosilicate, in general called as phosphor. It absorbs the kinetic energy of electrons striking it and re-emits in the form of light. Hence the screen is called as fluorescent screen. The colour of the trace on the screen depends on the type of material.

All these parts of CRT are enclosed in a funnel shaped highly evacuated glass envelope. The
inner surface of the flared part of this envelope is coated with an aqueous solution of graphite called aquadag, which is internally connected to the accelerating anode. This layer performs two functions-
(1) It accelerates the electron beam, (2) It collects the secondary electrons produced by secondary emission when electron beam strikes the screen.

(c) Exploits top down and bottom up approaches to prepare nanomaterials. [5M]

Engineered nano-objects and nanomaterials intended for industrial use may be synthesized using two different approaches known as the op down Approach and the Bottom Up Approach.

  1. Top-down approach
    The most common top-down approach to fabrication involves lithographic patterning techniques using short-wavelength optical sources. A key advantage of the top-down approach as developed in the fabrication of integrated circuits is that the parts are both patterned and built in place, so that no assembly step is needed. Optical lithography is a relatively mature field because of the high degree of refinement in microelectronic chip manufacturing, with current short-wavelength optical lithography techniques reaching dimensions just below 100 nanometers (the traditional threshold definition of the nanoscale).

Shorter-wavelength sources, such as extreme ultraviolet and X-ray, are being developed to allow lithographic printing techniques to reach dimensions from 10 to 100 nanometers. Scanning beam techniques such as electron-beam lithography provide patterns down to about 20 nanometers. Here the pattern is written by sweeping a finely focused electron beam across the surface. Focused ion beams are also used for direct processing and patterning of wafers, although with somewhat less resolution than in electron-beam lithography. Still-smaller features are obtained by using scanning probes to deposit or remove thin layers. Mechanical printing techniques—nanoscale imprinting, stamping, and molding—have been extended to the surprisingly small dimensions of about 20 to 40 nanometers.

  1. Bottom-up approach
    Bottom-up or self-assembly approach to nanofabrication use chemical or physical forces operating at the nanoscale to assemble basic units into larger structures. As component size decreases in nanofabrication, bottom-up approaches provide an increasingly important complement to top-down techniques. Inspiration for bottom-up approach comes from biological systems, where nature has harnessed chemical forces to create essentially all the structures needed by life. Researchers hope to replicate nature‘s ability to produce small clusters of specific atoms, which can then self-assemble into more-elaborate structures. A number of bottom-up approaches have been developed for producing nanoparticles, ranging from condensation of atomic vapors on surfaces to coalescence of atoms in liquids. For example, liquid-phase techniques have been developed to produce size-selected nanoparticles of semiconductor, magnetic, and other materials.

An example of self-assembly that achieves a limited degree of control over both formation and organization is the growth of quantum dots. Indium gallium arsenide (InGaAs) dots can be formed by growing thin layers of InGaAs on GaAs in such a manner that repulsive forces caused by compressive strain in the InGaAs layer results in the formation of isolated quantum dots. After the growth of multiple layer pairs, a fairly uniform spacing of the dots can be achieved. Another example of self-assembly of an intricate structure is the formation of carbon nanotubes under the right set of chemical and temperature conditions.

Image…….

Scroll to Top
error: Alert: Content selection is disabled!!