UD-TRG-Physics-Sem2-Ans-1paper

B.E. Second Semester All Branches (C.B.S.)

Time: 2 hours
Maximum marks: 60

Notes :

Question No. 1 is compulsory.
Attempt any three questions from the remaining questions.
Used suitable data wherever required.
Figures to the right indicate full mark.
Attempt any five: [15M]
(a) Why does an excessively thin film appear to be perfectly dark when illuminated by white light?

the expression for path difference in case of thin film is given by

\triangle=2\mu t\cos\theta\pm\frac\lambda2

If the film is excessively thin i.e. t << λ and the path difference Δ = λ/2. In this case condition for destructive interference is satisfied and therefore the film appears to be perfectly dark when illuminated by white light.

(b) A grating has 620 rulings/mm & is 5.05 mm wide. What is the smallest wavelength-interval that can be resolved in the third order at λ = 481 nm?

Since grating has 620 rulings/mm, number of rulings (N) in 5.05 mm = 620 x 5.05 = 3131

We have,

Resolving Power $=\frac\lambda{d\lambda}=Nn$ or

\begin{array}{l}d\lambda=\frac\lambda{Nn}\\\;\;\;\;\;=\frac{481\times10^{-9}}{3131\times3}\\\;\;\;\;\;=0.512\times10^{-10}\\\;\;\;\;\;=0.512\;\overset\circ A\\end{array}

Which is the smallest wavelength interval that can be resolved.

The reasons are as follows. Optical fibre are made from silica which is one of the most abundant material on the earth. The overall cost of optics communication is lower than that of an equivalent cable communication system. The cross section of an optical fibre is about few microns. Hence, the fibres are less bulky. Optical fibre is quite flexible and strong.

In optical fibres, information is carried by photons which are electrically neutral and cannot be disturbed by high voltage fields, lightening, etc. Therefore, fibres are immune to externally caused background noise generated through EMI and RFI.

The light waves propagation along the optical fibre are completely trapped within the fibre and cannot leak out further light cannot couple into the fibre from sides. Therefore, the possibility of cross talk is minimised when optical fibre is used. Thus, transmission is more secure and private.

Optical fibres have ability to carry large amounts of information. A 1mm fibre can transmit 50000 calls. The transmission loss per unit length of an optical fibre is about 1Db/km. Therefore, longer cables run between repeaters and are feasible.

(d) An electron is bound in a one dimensional potential well of width 2A0 but of infinite height. Find its energy values to the ground state and first excited state?

We have,

E=\frac{n^2h^2}{8mL^2}

For ground state, n = 1,

\begin{array}{l}E=\frac{\left(6.63\times10^{-34}\right)^2}{8\times9.1\times10^{-31}\times\left(2\times10^{-10}\right)^2}\\\;\;\;=0.3015\times10^{-17}\;J\\\;\;\;=18.8\;eV\end{array}

For first excited state, n = 2,

\begin{array}{l}E=\frac{25\times\left(6.63\times10^{-34}\right)^2}{8\times9.1\times10^{-31}\times\left(2\times10^{-10}\right)^2}\\\;\;\;=0.5375\times10^{-17}\;J\\\;\;\;=471\;eV\end{array}

(e) Explain measurement of frequency of AC signal using Cathode Ray Oscilloscope?

An AC signal whose frequency is to be measured is given to Y-input of CRO. The time-base sensitivity (time/div) is adjusted so as to obtain two or three cycles of AC on the CRO screen. The horizontal spread of one cycle is noted in terms of number of division. By multiplying it with time/div, the time for one cycle i.e. time period is obtained. The reciprocal of time period gives the frequency of the AC signal.

Image…….

(f) Explain the term Stimulated emission & Population inversion?

Stimulated emission:
A photon of energy hυ = E2 – E1 can induce an excited atom in higher energy state E2 to make downward transition to ground state E1, releasing excess energy in the form of photon. This phenomenon of forced emission of photon is called as induced emission or stimulated emission.

Image…..

The no. of stimulated transitions occurring in the material at any instant depends upon the no. of atoms at energy level E2 and spectral energy density. Thus, the number of atoms Nst that undergoes downward transition during the time Δt is

N_{st}\;/\Delta t\;\propto\;N_{2\;}Q

Hence, rate of spontaneous emission is,

\frac{dN_{st}}{dt}=B_{21}N_2Q

Where, B12 is the Einstein B- coefficient, N2 is the no. of atoms in state E2 and Q is the spectral energy density.

Population Inversion:
The no. of active atoms occupying an energy state is called as population of that state. The
population N of an energy state depends upon absolute temperature T and energy E.

In the state of thermal equilibrium there are more atoms in ground state than excited state.
However, in order to achieve stimulated emission, more no. of atoms should be in excited state than
ground state. Such a non-equilibrium condition in which number of atoms in excited state is greater than number of atoms in ground state is called as Population Inversion.

If N1 and N2 are the population of ground and excited state having energy E1 and E2 respectively
then population ratio is given by,

\frac{N_2}{N_1}=exp\left[\frac{\displaystyle-\left(E_2-E_1\right)}{kT}\right]

(g) Define superconductivity, critical temperature & critical magnetic field.

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.

Critical Field: 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.

Critical Magnetic Field: It is observed that superconductivity vanishes if a sufficiently strong magnetic field is applied. The minimum magnetic field which is necessary to regain the normal resistivity is called as critical magnetic field, Hc.

(a) How is Newton’s ring experiment used to determine refractive index of liquid medium? The diameter of 5th dark ring in Newton’s ring experiment was found to be 0.42 cm. Determine the diameter of 10th dark ring [8M]

With air as a medium, let Dn and Dn+p be the diameters of nth and n+pth dark ring respectively, where p is any number. Then we can write

D_n^2=4n\lambda R\;and\;D_{n+p}^2=4\left(n+p\right)\lambda R

\begin{array}{l}\therefore D_{n+p}^2-D_n^2\\\;\;\;=4\left(n+p\right)\lambda R-4n\lambda R\\\;\;\;=4p\lambda R\end{array}

If the gap between the lens and plane glass sheet is filled with a liquid, the air film is substituted by the liquid film. The condition for interference is now given by

2\mu t\;\cos r=n\lambda

For normal incidence,

2\mu t=n\lambda

Where μ is the refractive index of the liquid. The radius of the nth dark ring is given by,

\begin{array}{l}r_n^2=2Rt\\{\left(r_n^2\right)}_L=\frac{n\lambda R}\mu\\{\left(Dn^2\right)}_L=\frac{4n\lambda R}\mu\end{array}

Similarly, the diameter of (n+p)th ring is given by

\begin{array}{l}\left(D_{n+p}^2\right)=\frac{4\left(n+p\right)\lambda R}\mu\\{\left(D_{n+p}^2\right)}<em>L-{\left(Dn^2\right)}_L=\frac{4p\lambda R}\mu\\\mu=\frac{{\left(D</em>{n+p}^2\right)}<em>{air}-{\left(Dn^2\right)}</em>{air}}{{\left(D_{n+p}^2\right)}_L-{\left(Dn^2\right)}_L}\end{array}

Measuring the values of Diameters of the Newton’s ring in air and liquid the refractive index can be of the liquid can be calculated by using the above formula.

The diameter of dark ring is given by,

D_n\;=\;\sqrt{4n\lambda R}

Therefore, $D_5=\sqrt{4\times5\lambda R}\;\;and\;\;D_{10}=\sqrt{4\times10\lambda R}$

Thus,

\begin{array}{l}\frac{D_{10}}{D_5}=\frac{\sqrt{10}}{\sqrt5}=\sqrt2\;or\\D_{10}=1.414\times D_5\\\;\;\;\;\;\;\;=1.414\times0.42\\\;\;\;\;\;\;\;=0.59\;cm\end{array}

(b) An optical fibre has core diameter of 6μ and its core refractive index 1.45. The critical angle is 87°. Calculate- (i) Refractive index of cladding (ii) acceptance angle (iii) The number of modes propagating through fibre when wavelength of light is 1 μm. [7M]

i) We have,

\begin{array}{l}\sin\theta_c=\frac{n_2}{n_1}\;or\;n_2=n_1\;\\\sin\theta_c=1.47\times\sin\;87^\circ\\\;\;\;\;\;\;\;\;\;=1.467\end{array}

ii) We Have,

\begin{array}{l}\theta_0=\sin^{-1}\left(\sqrt{n_1^2-n_2^2}\right)\\\;\;\;\;=\sin^{-1}\left(\sqrt{1.47^2}-1.467^2\right)\\\;\;\;\;=5.38\;degree\end{array}

iii) We have,

\begin{array}{l}v=\frac{\pi d}\lambda NA\\\;\;\;=\frac{3.14\times66\times10^{-6}\times\sin\;5.38}{1\times10^{-6}}\\\;\;\;=1.912<2.405\end{array}

Hence the fiber is single mode step index fiber i.e. only one mode can propagate through the fiber.

(a) With neat energy level diagram, explain principle, construction & working of Nd-YAG laster? [8M]

Nd-YAG laser is a four level solid state laser. The active medium is YAG (Yttrium Aluminium Garnet, Y3Al5O12) crystal wherein some Y3+ ions are replaced by Nd3+ ions, which acts as active centres.

Construction: The basic construction of Nd-YAG laser is shown in figure. It consists of an cylindrical
reflector housing the laser rod along one side of its focus lines and a flash lamp along the other focus line. The light leaving one focus of the ellipse passes through the other focus after reflection from the silvered surface of the reflector. Thus, the entire flash lamp radiation gets focused on the laser rod. The two ends of the rods are gets polished and silvered and constitute optical resonator.

Image…….

Working: The energy level of Nd ion in YAG crystal is shown in figure. The energy level structure of Nd atom is preserved to a certain extent because of its relatively low concentration. However the energy levels are split and the structure is complex. The pumping of the Nd ions to the upper states is done by krypton arc lamp. The optical pumping with light of wavelength 5000 to 8000 Å excites the ground state Nd ions to higher states.

The metastable state E3 is upper lasing level whereas E2 forms the lower lasing level. The upper level E3 will be rapidly populated as the excited Nd ions quickly make downward transition from the upper energy bands. The lower energy level E2 is far above the ground level and hence it cannot be populated by Nd ions through the thermal transitions from the ground state. Therefore, the population inversion is readily achieved between the E2 and E3 levels. The laser transition occurs in the infrared (IR) region at a wavelength of about 10600 Å. As the laser is four level laser the population inversion can be maintained for continuous laser emission.

Thus, Nd-YAG laser can be operated in CW mode with efficiency better than 1%.

Image….

(b)Two plane rectangular pieces of glass are in contact at one edge & are separated at the other end 10 cm away by a wire to form a wedge shaped film.When the film was illuminated by light of wavelength 6000A°, 10 fringes were observed per cm. Determine the diameter of the wire. [7M]

Given,

l=10\;cm,\;\lambda=6\times10^{-5}cm,\;p=10,\;x_{n+p}-x_n=1\;cm,\;d=?

We have, diameter of wire,

\begin{array}{l}d=t\\t=\frac{lp\lambda}{2\left(x_{n+p}-x_n\right)}\\\;\;\;=\frac{10\times10\times\times6\times10^{-5}}{2\times1}\\\;\;\;=3\times10^{-3}cm\\\;\;\;=0.03\;mm\end{array}

\begin{array}{l}d=t\\L\theta=\frac{100\beta\times\lambda}{2\beta}\\\;\;\;\;\;=50\lambda\\\;\;\;\;\;=50\times6000\;\overset\circ A\\\;\;\;\;\;=30,0000\;\overset\circ A\\\;\;\;\;\;=3\times10^{-5}m\\\;\;\;\;\;=0.03\;mm\end{array}

(a) Explain the experimental method to determine the wavelength of spectral line using diffraction grating? [5M]

In order to determine the wavelength of different spectral lines of sodium using plane diffraction
grating. First adjust the spectrometer for parallel rays by Schuster’s method and the grating is adjusted for normal incidence as follows:Bring the telescope in the line of the collimator to have the image of the slit on vertical cross wire without the grating on the prism table and note down the reading. Now rotate the telescope through 90° in one direction and clamp it. Now place the grating on the prism table and rotate the table to get reflected image of the slit on the vertical cross wire of the telescope. This means the angle of incidence is now 45°.
Note the position of prism table and rotate the prism table through 45°, so that light falls normally on the grating. Fix the prism table.

Now rotate the telescope to the left side and adjust it to get two closely spaced yellow images of
the slit. These are the two lines, D1 and D2, as observed in the first order diffraction spectrum for which n = 1.

Set the vertical cross wire on one of the lines and take the readings in both the windows, W1 and
W2. Take the readings for the other line by slowly shifting and setting the cross wire on it.

Rotate the telescope in the same direction so as to get D1 and D2 again in the field of view. This is the second order diffraction spectrum corresponding to the case n = 2. This spectrum shows a larger separation between D1 and D2 and smaller intensities of these lines compared to that in the first order. Take the readings for D1 and D2 as explained as above.

Now rotate the telescope to the right side, adjust it and take readings for the first and the second order spectra as explained above.

Image…..

Tabulate the observations in the observation table and obtain the angles θ1 and θ2 for the first and
the second order spectra by taking the difference between the readings for the left and the right side.

The wavelength of different spectral lines of sodium is given by

\lambda=\frac{\left(a+b\right)\sin\theta n}n

Putting values of grating element (a+b) and θ1 and θ2 the wavelengths of the spectral lines are is
calculated.

(b)Show that electron cannot pre-exist in free state in a nucleus, using uncertainty principle. [5M]

If electron is inside the nucleus, we will get it within its diameter. i.e. $\Delta x=\;2\;\times10^{-14}\;m.$

Now,

\Delta x\;\Delta p\;=\frac{\;h}{2\pi}\;\;\boldsymbol o\boldsymbol r\;\;\Delta p\;\geq\frac{\;h}{2\pi}\;\Delta x\;\approx\;5\;x\;10-21\;kgm/s.

This, minimum energy, $E=\;p.c\;=\;mc.c$

\begin{array}{l}=\;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.

(c) Distinguish between type I & type II superconductor? [5M]
Type –I Superconductor Type –II Superconductor
Transition from superconducting state to
normal state is sharp Transition from superconducting state to normal
state occurs gradually, as it has two critical field
Perfectly diamagnetic below Hc, and
completely expels the magnetic field Is in mixed magnetic state between two critical
fields
Poor carriers of electric current Can carry large magnetic current
The critical field is relatively low The critical magnetic field is relatively high
Examples are Al, Pb, In etc Examples are transition metals and alloys etc
5(a)A diffraction grating used at normal incidence gives a yellow line (λ = 6000A°) in a certain spectral order superimposed on a blue line (λ = 4800A°) of next higher order if the angle of diffraction is sin-1(3/4), calculate the grating element? [5M]

We have,

\begin{array}{l}\left(a+b\right)\;\sin\theta=n\lambda_{yellow}\\\left(a+b\right)\;\sin\theta=\left(n+1\right)\lambda_{blue}\end{array}

Dividing

\begin{array}{l}1=\frac{n\lambda_{yellow}}{\left(n+1\right)\lambda_{blue}}\\n6000=\left(n+1\right)4800\\n6=4.8n+4.8\\or\\n=4\end{array}

From first equation,

\begin{array}{l}=(a+b)\;x\;3/4\;\\=\;4\;x\;6\;x\;10-5\;cm\;\\=\;32\;x\;10-5\;cm\end{array}

(b)Derive one dimensional time dependent Schrodinger’s equation for matter waves? [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)With neat diagram, explain construction & 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.

(a) Find the de broglie wavelength of (i) an electron accelerated through a potential difference of 182 volts &
(ii) 1 kg object moving with a speed of 1 m/s. Comparing the results, explain why is the wave nature of matter not apparent in daily observations? [5M]

We have,

i) de Broglie wavelength of electron,

\begin{array}{l}\lambda=\frac{12.26}{\sqrt V}\\\;\;\;=\frac{12.26}{\sqrt{182}}\\\;\;\;=0.9087\;\overset\circ A\end{array}

ii) de Broglie wavelength of object,

\begin{array}{l}\lambda=\frac h{mv}\\\;\;\;=\frac{6.63\times10^{-34}}{1\times1}\\\;\;\;=6.63\times10^{-24}\;\overset\circ A\end{array}

As is clear from the results the de-Broglie wavelength of an electron is of the order of wavelength of Xray and can be measureable. Whereas the wavelength of a 1 kg object is so small to detect by any of the method. This is the reason, why the wave nature of matter is not apperent in our daily observation.

(b) Derive Bethe’s law for electron refraction? [5M]

Image……..

A non uniform field is field in which electric intensity varies from point to point. Non – uniform
electric fields are represented by equipotential surfaces. On an equipotential surface, the electric potential remains constant and the electric field lines are normal to the surface at any point.

When electron travels through a region of uniform potential V1 with a velocity v1 into a region of uniform potential V2, it experiences a force which changes it’s velocity and beam of electron can bend from it’s original path. As electric field exists only in the vertical direction, the normal component of the electron velocity Vy undergoes a change whereas the tangential component Vx remains constant. Bethe’s law gives the bending of an electron beam at an interface between two regions with different potentials whereas the Snell’s law gives the bending of light rays at an interface between two media with different refractive indices.

Thus, Snell’s law for refraction of light is similar or analogous to the Bethe’s law for refraction of electron beam.

Let XY be the surface of separation of two regions.
V1 and V2 are the potentials in regions A and B respectively with V2 greater than V1. Let v1 and v2 are velocities of electrons in regions A and B respectively.
θ1 and θ2 be the angles of incidence and angle of refraction of electron beam respectively.
If ‘e’ and ‘m’ are the charge and mass of an electron respectively. Then kinetic energy for region A is

\left(\frac12\right)mv_1\;=\;eV_1

v_1=\left{\frac{2eV_1}m\right}^{1/2}-----\left(i\right)

Similarly for region B,

v_2=\left{\frac{2eV_2}m\right}^{1/2}-----\left(ii\right)

Dividing equation (i) and (ii),

\frac{v_1}{v_2}=\left{\frac{V_1}{V_2}\right}^\frac12-----\left(iii\right)

If these velocities v1 and v2 are resolved into two components parallel and perpendicular to the surface XY, then v1cos θ1 and v2cos θ2 are their perpendicular components and v1sin θ1 and
v2sin θ2 are their parallel components. XY is the equipotential surface and hence there is no change in the component parallel to XY.

Hence, v1sin θ1 = v2sin θ2

\frac{v_1}{v_2}=\;\frac{\sin\;\theta_1}{\sin\;\theta_2}\;-----\left(iv\right)

or by using equation number (iii), we can write equation no. (iv) as,

\frac{\sin\theta_1}{sin\theta_2}=\frac{v_1}{v2}=\left{\frac{V_1}{V_2}\right}^\frac12

When V2 is greater than V1, the beam is deflected towards the normal and when V2 smaller than V1, thebeam is deflected away from the normal. This is called as Bethe’s law.

Carbon nanotube (CNT) structures consist of a graphene sheet(s) rolled into a tubular array. Based
on the layers of graphene sheets that comprise the CNT, the structures are designated as single-walled nanotubes (SWNTs), double-walled nanotubes (DWNTs), or multi-walled nanotubes (MWNTs).

The production of displays with low energy consumption could be accomplished using carbon
nanotubes (CNT). Carbon nanotubes are electrically conductive and due to their small diameter of several nanometers, they can be used as field emitters with extremely high efficiency for field emission displays (FED). The principle of operation resembles that of the cathode ray tube, but on a much smaller length scale.