Jan-2020- Solution

January-2020

Engineering Physics (18PHY12/22)

Time- 3 Hours

Maximum Marks- 100

Module – 1

Q.1. a) Give the theory of forced vibrations and obtain the expression for amplitude. [08 M}

Ans. When a continuous and periodic external force is applied to the body, then the vibrations of the body are said to be forced vibrations. When there is steady state, the body vibrates with a frequency of external periodic force.

When the body oscillates, it is subjected to three forces :

i) Restoring force having magnitude proportional to displacement from mean position and is always directed towards the mean position. It is always written as -μy where μ is the force constant and y is the displacement.

ii) Frictional force having magnitude proportional to velocity and is directed opposite to it. It is written as -r \frac{dy}{dt} where r is known as damping constant or damping coefficient constant.

iii) The external periodic force of angular frequency p and maximum magnitude F. It can be written as F sin pt.

F_net= -μy-rdy/dt+ Fsin pt

F_{net}= m\frac{d^{2}y}{dt^{2}} m\frac{d^{2}y}{dt^{2}}= -\mu y-r\frac{dy}{dt}+F\; sin pt m\frac{d^{2}y}{dt^{2}}+\mu y+r\frac{dy}{dt} = F\; sin pt \frac{d^{2}y}{dt^{2}}+\frac{\mu }{m} y+\frac{r}{m}\frac{dy}{dt} = \frac{F}{m}\; sin pt

Let,  \frac{r}{m}=2b, \frac{\mu }{m}=\omega ^{2} and\; \frac{F}{m}=f

\frac{d^{2}y}{dt^{2}}+\omega ^{2} y+2b\frac{dy}{dt} = f\; sin pt—————(1)

The above equation is the differential equation for forced SHM.

The genera solution of this equation is of the type

y=A sin(pt-θ)————–(2)

Where θ is the phase angle through which the displacement lags the applied force and A is the amplitude.

Differentiating the above equation twice with respect to t,

\frac{dy}{dt}= A p cos(pt-\theta ) \frac{dy^{2}}{dt^{2}}= -A p^{2} sin(pt-\theta )

Substituting in equation (1)

-Ap² sin (pt-θ)+2b A p cos (pt-θ) +ω²A sin(pt-θ)=f sin pt

-Ap² sin (pt-θ)+2b A p cos (pt-θ) +ω²A sin(pt-θ)=f sin [(pt-θ) cos θ +f cos (pt-θ) sin θ

A (ω²–p²) sin (pt-θ) +2b Apcos(pt-θ )= f sin (pt-θ) cos θ +f cos(pt-θ) sin θ

For the above equation to be true for all values of time t, the coefficients of sin(pt-θ) and cos(pt-θ) must be equal on the two sides.

A(ω²–p²)=f cosθ ————————–(3)

and 2b Ap=f sin θ—————————–(4)

Squaring and adding equations (3) and (4), we get

A(ω²–p²)²+ 4b²A²p²=f²

A²[(ω²–p²)²+4b²p²]=f²

A=\frac{1}{\sqrt{[(\omega ^{2}-p^{2})^{2}+4b^{2}p^{2}]}}———————————(5)

Dividing equation (4) by equation (3),

tan \theta =\frac{2bp}{w^{2}-p^{2}}

\theta = tan^{-1}\left ( \frac{2bp}{w^{2}-p^{2}} \right ) ———————(6)

1. If p<<ω, the amplitude of vibration given by equation (5) becomes

A=\frac{f}{\omega ^{2}}=\frac{\left ( \frac{F}{m} \right )}{\left ( \frac{\mu }{m} \right )}=\frac{F}{\mu }= constant

and from equation (6),

θ=tan^-1 0=0

Hence, if p<<ω, amplitude of vibration becomes independent of frequency of force. It depends only on the magnitude of applied force and the force constant.

Also, the force and the displacement remain in phase.

2. If p=ω, the amplitude of vibration given by equation (5) becomes,

A=\frac{f}{2bp}=\frac{\left ( \frac{F}{m} \right )}{\left ( \frac{r}{m} \right )p}=\frac{F}{rp} \theta =tan^{-1}\left ( \frac{bp}{0} \right )=\frac{\pi }{2}

For small damping(i.e., small r), the amplitude of vibration will be large. The displacement lags behind the force by π/2.

3. If p>>ω, the amplitude of vibration given by equation (5) becomes

A=\frac{f}{\sqrt{[(p^{2})+4b^{2}p^{2}]}}\approx \frac{f}{p^{2}}=\frac{F}{mp^{2}} \theta =tan^{-1}\left ( \frac{2bp}{-p^{2}} \right )=tan^{-1}\left ( \frac{-2b}{p} \right )\approx tan^{-1}(-0)=\pi

As p is large, amplitude of oscillation will be small. The displacement lags behind the force by π.

This is an expression of amplitude.

b) With a neat diagram, explain the construction and working of Reddy tube. Mention four applications of shock waves. [08 M]

Ans. Reddy’s shock tube is a hand operated device which can produce shock waves with Mach number greater than 1.5.

Construction- i) It consists of a steel tube of length about 1 m and a diameter of about 30 mm.

ii) The tube is divided into two sections by a 0.1 mm thick diaphragm made aluminium or mylar, each about 50 cm in length as shown in Fig.

iii) On section is known as the driver section and the other is called driven section.

iv) The two sections generally contain two different gases known as the driver gas and the test gas. The end of the tube in the driver section is provided with a piston and the other end is closed.

v) A pressure gauge is attached near the diaphragm and two pressure sensors are mounted near the closed end.

Working- i) The piston is pushed hard due to which the pressure in the driver section increases and the diaphragm ruptures.

ii) The driver gas enters the driven section suddenly creating a shock wave which travels towards the closed end. This increases the pressure and temperature of the test gas.

iii) When there is  reflection from the closed end, the reflected shock wave further increases the pressure and temperature of the test gas.

iv) When the diaphragm is ruptured, a compression pulse travels into the test gas and an expansion pulse travels in opposite direction. The expansion pulse gets reflected from the piston, travels back and partially neutralizes the reflected compression pulse.

v) The two sensors p₁ and p₂ sense the pressure changes in the primary and the reflected shock waves. The time interval between the pulses can be obtained by feeding the output of the sensors to a CRO.

vi) If the distance between the two sensors i.e d and the time interval between the pulses is t, the speed of the shock waves can be obtained using v=d/t

Knowing the speed of sound in the test gas, the Mach number can be calculated.

Applications of shock waves- 1. They are used in the treatment of kidney stones.

2. Shock waves are used to treat fractures as they activate the healing process in tendons and bones.

3. Shock waves are used in metallurgy for compaction of ceramic powders.

4. The shock waves are used in wave disc engines to transfer energy from a fluid having large energy to a fluid having small energy.

5. Shock waves develop when objects like jets and rockets move at supersonic speeds. Hence the shock waves are studied to develop designs for jets, rockets and high speed turbines.

6. Shock waves develop in pipes through which fluids flow at supersonic speeds. Hence they are studied to develop designs for ducts and pipes.

c) Calculate the resonant frequency for a simple pendulum of length 1 m. [04 M]

Ans. Solution- n=\frac{1}{2\pi }\sqrt{\frac{3}{l}}

g=9.81 m/s², l=1m

n=\frac{1}{2\pi }\sqrt{\frac{9.81}{1}}

n=0.5 Hz

OR

Q.2. a) Define force constant and mention its physical significance. Derive the expression for force constant for springs in series and parallel combination.[08 M]

Ans. i) Consider a light vertical spring of length L hanging from a rigid support as shown in Fig,.

When small mass m is attached at the lower end, the length of the spring increases by l under equillibrium condition as shown in Fig.

According to Hooke’s law, if the elongation l of the spring is within its elastic limit, the restoring force (T) inside it is proportional to the elongation and tries to restore the original length of the spring.

T∝ l

T=kl

where, k is the constant of proportionality known as the force constant.

ii) The force constant represents the force developed in the spring per unit length. It represents the stiffness of the spring. Hence it is also known as the stiffness constant of the spring.. As the mass is in equillibrium.

T=kl=mg

iii) If the mass is now given a small displacement y in the downward direction as shown in Fig  and released, the total elongation of the spring becomes l+y. Hence, the force in the spring is k(l+y) which acts in the upward direction.

iv) The net force on the block is

mg-k(l+y)=mg-kl-ky

As kl=mg, net force on block is

mg-mg-ky=-ky

The net force equals.

Therefore, ma = -ky

a=-\frac{k}{m}y

As, a=-ω²y,

The time period is

T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}

The frequency is

n=\frac{1}{T}=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{m}}

a) Expression for force constant for springs in series

Consider two springs in series . Let k₁ and k₂ be the two force constants connected in series. They both are connected to the mass m.

When the springs are stretched there is an elongation in the springs. So let the elongation in springs be x₁ and x₂. The total displacement of the mass will be

x=x₁+x₂ ———-(1)

These two springs have same axial force.

mg=k₁x₁=k₂x₂

x_{1}=\frac{mg}{k_{1}}

and x_{2}=\frac{mg}{k_{2}}

Considering a complete system of springs,

mg=k_sx

x=\frac{mg}{k_{s}}

Putting these values in equation (1)

\frac{mg}{k_{s}}=\frac{mg}{k_{1}}+\frac{mg}{k_{2}}

\frac{1}{k_{s}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}——————————————(2)

k_{s}=\frac{k_{1}k_{2}}}{k_{1}+k_{2}}

For more than two springs we can write series combination as follows.

\frac{1}{k_{s}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}+---

b) Expression for springs in  parallel combination of force constant

Consider the two springs connected in parallel. Let k1 and k2 be the two force constants connected in parallel with the spring, connectde to mass m.

Let x be the displacement of the mass. The elongation of springs be x. The two springs are connected in parallel so they have different axial forces.

The forces in the two springs are given by,

F₁=k₁x

F₂=k₂x

For equllibrium of the block,

F=F₁+F₂=mg

mg=k₁x+k₂x

If k_p is the equivalent force constant,

mg=k_px

k_px=k₁x+k₂x

k_p=k₁+k₂

For more than two springs,

k_p=k₁+k₂+k₃+——

These is the expression for force constant connected in parallel.

These are the expression for series and parallel combination of force constants.

b) Define simple harmonic motion. Derive the differential equation of motion for it using Hook’s law. Mention the characteristics and examples of simple harmonic motion. [08 M]

Ans. Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position.

According to Hooke’s law, the restoring force is directly proportional to the displacement (y) of the mass from mean position and is always directed towards the mean position.

F ∝ y

F=-ky

Where, k is called constant of proportionality and the negative sign shows that the force is opposite to the displacement.

We know that, F=ma

Where m is mass of the body  which performs simple harmonic motion and a is acceleration of the body.

The acceleration a is written as

a=\frac{d^{2}y}{dt^{2}} m\frac{d^{2}y}{dt^{2}}=-ky

The above equation differential form can be written as

\frac{d^{2}y}{dt^{2}}+\left ( \frac{k}{m} \right )y=0

This is differential equation of motion using Hooke’s law.

The characteristics of Simple Harmonic motion are as follows :-

1. Displacement- Displacement is a vector quantity. Its magnitude is equal to the distance of the body performing SHM from the mean position and its direction is represented by a positive or negative sign indicates the direction of the body towards the mean position.
2. Amplitude- The magnitude of the maximum displacement of the body is called as the amplitude.
3. Velocity- The velocity of the body at any time t is given by v=\frac{dy}{dt}= A\omega cos(wt+\phi ) at any dispalcemnet y, it is given by , v=\omega \sqrt{A^{2}-y^{2}} . At extreme positions the velocity of the body is zero and has the maximum magnitude Aω at the mean position.
4. Acceleration- The acceleration is proportional to the displacement from mean position and is directed opposite to the displacement. It is given by a=\frac{dv}{dt}=-A\omega ^{2}sin(\omega t+\phi )=-\omega ^{2}y . The acceleration is zero at mean position and has a maximum magnitude of ω²A at the extreme positions.
5. Time period- The time taken by body to complete one oscillation. It is given by, T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}
6. Frequency- The number of oscillations per second is given by n=\frac{1}{T}=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{m}}
7. Phase- The state of vibration is given by the phase. i.e displacement of the body and the direction of the motion. The quantity ωt-Φ is phase of SHM. Φ is the initial phase or epoch.

Examples of SHM- 1. A cantilever beam of negligible mass with small mass attached to its free end and when given a small transverse displacement it performs a SHM.

2. When a body floats in a liquid it is given vertical displacement then it performs SHM.

c) The distance between the two pressure sensors in a shock tube is 150 mm. The time taken by a shock wave to travel this distance is 0.3 ms. If the velocity of sound under the same condition is 340 m/s. Find the mach number of the shock wave. [04 M]

Ans. v=d/t

t=0.3 ms= 0.3 × 10^-3 s, d=150 mm=150 × 10^-3 m

v=\frac{150\times 10^{-3}}{0.3\times 10^{-3}}=500 m/s M=\frac{v}{a}

a=340 m/s

M=\frac{500}{340}=1.47

M=1.47

Module-2

Q.3. a) Explain longitudinal stress, longitudinal strain, volume stress and volume strain. Discuss the effect of stress temperature, annealing and impurities on elasticity.[08 M]

Ans. When any body is experiencing an external force on it, the force causes deformation in it. The internal resisting forces are developed inside the body and these forces try to restore the original shape and size of the body. The original shape and size gets changed. These internal forces which try to restore the original shape and size of the body are called as restoring forces.

If deformation is within elastic limit , hen the magnitude of restoring force is equal to magnitude of deforming force.

The restoring force set up in  a body subjected to deforming forces per unit cross sectional area is known as  stress.

Let R= resisting force and

A= Cross sectional area,

Stress= R/A

The S.I unit of stress is pascal (Pa) i.e equivalent to N/m2.

When a wire or rod is subjected to forces it try to elongate or compress them, the stress  developed in them is known as longitudinal stress. Longitudinal stress can be either tensile or compressive.

b) Derive the relation between bulk modulus (k), Young’s modulus (Y) and Poisson’s ratio (α), What are the limitting values of Poisson’s ratio? [08 M]

c) Calculate the extension produced in a wire of length 2m and radius 0.013 × 10^-2 m due to a force of 14.7 Newton applied along its length. Given, Young’s modulus of the material of the wire Y= 2.1 × 10¹¹ N/m²

Q.4. a) Describe a single cantilever and derive the expression for Young’s modulus of the material of rectangular beam. [08 M]

b) Derive an expression for couple per unit twist for a solid cylinder with a diagram. [08 M]

c) Calculate the angular twist of a wire of length 0.3 m and radius 0.2 × 10 ^-3 m when a torque of 5 ×10^-4 Nm is applied. (Rigidity modulus of the material is 8 × 10¹⁰ N/m²). [04 M]

Module-3

Q.5. a) explain divergence and curl. Derive Gauss divergence theorem. [08 M]

Ans.i)  The Divergence of a vector is defined as is defined as the partial derivative of P with respect to x plus the partial derivative of Q with respect to y plus the partial derivative of R with respect to z.

div F= \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} Div\overrightarrow{F}=\overrightarrow{\bigtriangledown}.\overrightarrow{F}

= \left ( \widehat{i}\frac{\partial }{\partial x} +\widehat{j}\frac{\partial }{\partial y}+\widehat{k}\frac{\partial }{\partial z}\right )= {F_{x} ( \widehat{i})+F_{y} ( \widehat{j})+F_{z} ( \widehat{k})}

\overrightarrow{\bigtriangledown }\cdot \overrightarrow{F}=\left (\frac{\partial F_{x}}{\partial x} +\frac{\partial F_{y} }{\partial y}+\frac{\partial F_{z}}{\partial z}\right )

Thus, \overrightarrow{\bigtriangledown }\cdot \overrightarrow{F} is a scalar quantity.

ii) Curl of a vector– The curl of a vector is defined as the vector operation of differential vector \overrightarrow{\bigtriangledown} on vector F.

\overrightarrow{\bigtriangledown }\times \overrightarrow{F} = \left ( \widehat{i}\frac{\partial }{\partial x} +\widehat{j}\frac{\partial }{\partial y}+\widehat{k}\frac{\partial }{\partial z}\right )= {F_{x} ( \widehat{i})+F_{y} ( \widehat{j})+F_{z} ( \widehat{k})}

\overrightarrow{\bigtriangledown }\times \overrightarrow{F}=\begin{bmatrix}\widehat{i} & \widehat{j} & \widehat{k} \\\frac{\partial }{\partial x} &\frac{\partial }{\partial x} & \frac{\partial }{\partial x}\\F_{x}& F_{y} & F_{z}\end{bmatrix} \widehat{i}\left ( \frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z} \right )+\widehat{j}\left ( \frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x} \right )+\widehat{k}\left ( \frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y} \right )

iii) The curl of a field is formally defined as the circulation density at each point of the field. A vector field whose curl is zero is called irrotational.

Gauss Divergence Theorem-The divergence theorem states that the surface integral of the normal component of a vector point function “F” over a closed surface “S” is equal to the volume integral of the divergence of vector F   taken over the volume “V” enclosed by the surface S.. i.e. \oint_{s}\overrightarrow{F}.\overrightarrow{ds}=\int _{V}(\overrightarrow{\bigtriangledown }.\overrightarrow{F})dV

Proof- i) Consider the differential elements with sides dx, dy and dz parallel to x, y and z axis as shown in fig.

ii) Let \overrightarrow{F} be the vector field at O. The flux of \overrightarrow{F}  diverging from the element =

\overrightarrow{F}.ds———————–(i)

iii) The x-component of vector field at face is

AEHD = F_{x}-\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx

Flux through AEHD = F_{x}-\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx dy dz

The x- component of vector field at face BCGF= F_{x}+\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx

Flux through BCGF = F_{x}+\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx dy dz

iv) The Net flux through the two faces parallel to y-z plane = F_{x}+\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx dy dz- F_{x}-\frac{1}{2}\frac{\partial F_{x}}{\partial x}dx dy dz

= \frac{\partial F_{x}}{\partial x}\; dx \; dy\; dz

Similarly, The net flux through the two faces parallel to x-z plane is = \frac{\partial F_{y}}{\partial x}\; dx \; dy\; dz and net flux through the two fac4es parallel to x-y plane = \frac{\partial F_{z}}{\partial x}\; dx \; dy\; dz

v) The total outward flux from differential element = \frac{\partial F_{x}}{\partial x}\; dx \; dy\; dz+= \frac{\partial F_{y}}{\partial x}\; dx \; dy\; dz+ \frac{\partial F_{z}}{\partial x}\; dx \; dy\; dz

\left ( \frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z} \right ) dx\; dy\; dz = \left ( \overrightarrow{\bigtriangledown }.\overrightarrow{F} \right )dV—————————-(ii)

vi) From eq. (i)  the flux through a larger surface S will be  \oint \overrightarrow{F}. \overrightarrow{ds}

vii) From eq. (ii) The total flux through surface S will be \int _{V}\left ( \overrightarrow{\bigtriangledown }.\overrightarrow{F }\right )dV

Therefore,   \oint \overrightarrow{F}. \overrightarrow{ds} = \int _{V}\left ( \overrightarrow{\bigtriangledown }.\overrightarrow{F }\right )dV

Hence, the Gauss Divergence Theorem is proved.

b) Define V-number and fractional index change. With a neat diagram, explain different types of optical fibres. [08 M]

Ans. i) V-Number- The V number is a dimensionless parameter which is often used in the context of step-index fibers. It is defined as. where λ is the vacuum wavelength, d is the radius of the fiber core, and NA is the numerical aperture. It is also called normalized frequency. The no. of mods that can travel in a fibre can be determined from V-number which is given by,

V=\frac{\pi d}{\lambda }\times (N.A)

where, λ is the wavelength of light used and d is the diameter of core of the optic fibre.

ii) Fractional index change- The factional index change  which is also known by Δ is also called relative refractive index is defined as the ratio of difference between the core and cladding to the refractive index of the core.

\bigtriangleup =\frac{n_{1}-n_{2}}{n_{1}}

N.A= Sin θ₀= \sqrt{n_{1}^{2}-n_{2}^{2}}

N.A= \sqrt{(n_{1}+n_{2})(n_{1}-n_{2})}

For small difference between n₁ and n₂,

n₁+n₂≈2n₁

n₁+n₂=n1Δ

N.A=\sqrt{(2n_{1})(n_{1}\bigtriangleup )} N.A=n_{1}\sqrt{2\Delta }

Thus, the numerical aperture can be increased by increasing fractional index change.

Types of optical fibres- According to the following criteria, like i) Material used ii) No. of modes and, iii) Variation of refractive index along the diameter. optical fibres  have following types :

1. Glass and plastic fibres – i) Glass optic fiber consists of glass core covered by a glass cladding having smaller refractive index.

ii) As compared to plastic fibres, the glass fibres have lower attenuation and large mechanical strength. Silica (SiO₂) is the most commonly used material  in glass fibre. It can be doped with aluminium oxide to increase its refractive index or with fluorine to decrease the refractive index.

iii) Silica fibres can be drawn into thinner fibres as compared to plastic fibres and hence they are suitable for long distance communications.

iv) Plastic fibres has core of polystyrene and cladding of poly-methyl methacrylate ( commonly known as PMMA). Plastic fibres are more flexible as compared to glass fibres. They have more attenuation and they can’t be drawn into thin fibres. Hence, they are suitable for short distance communications.

2. Single Mode and Multi Mode fibres –

i) When light is incident on optical fibre cores having diameter 50μm to 200 μm and it travels along different paths which is called modes of propagation.

ii) Ray which travels along axis is called axial mode and higher order modes are rays which travels at smaller angles of incidence on core-cladding interface. For small angle of incidence light does not transmit through optical fibre. The higher order mode, ray is incident on core-cladding interface.. The fig shows different modes of transmission. This fibres ar called multi mode fibres.

iii) As compared to higher order mode , the distance travelled by axial mode is small . This is the reason that the different modes reach other end at different times. This si called intermodal dispersion. This feature is undesirable because it causes signal distortion. It can only be eliminated by reducing no. of modes travelling through fibres and by reducing the diameter of core having small core diameters less than 10 μm and it transmits a single mode . Due to this intermodal dispersion is eliminated. Such fibres are known as single mode fibres.

3. Step index and Graded Index fibres –

Step Index fibres – i) The fibres which have core of transparent material of refractive index n, and cladding of homogenous transparent material of refractive index n₂ (<n₁). Due to changes in core -cladding interface there is abrupt change in refractive index. Therefore, they are called step index fibres. It has core diameter of order 50 μm to 200 μm. It transmits large number of modes as shown in fig.

ii) The thickness of cladding is 20 to 25 μm . This fibres are called multimode step index fibres. As compared to other fibres, these fibres are cheaper but they are not suitable for long distance communications . Intermodal dispersion causes significant signal distortion.

iii) The step index fibres having core diameters in  range 2 μm to 10 μm is called single mode step index fibres. These fibres transmits only axial mode as they have small core diameter. Due to elimination of intermodal dispersion these fibres are used for long distance communications.

Graded Index (GRIN) multimode fibres- i) In these the refractive index is maximum at axis of core and it decreases gradually upto cladding and then it remains constant throughout the cladding. Due to gradually decreasing refractive index the rays travels at an angle to axis travelling  in curved paths.

ii) As compared to axial rays these rays travels long distances and they travel in region of small refractive index and hence they travel faster.  Due to this intermodal dispersion is reduced. The core diameter ranges from 20 μm to 100 μm and the thickness of cladding is about 25 μm. These fibres are used for medium distance communications.

c) Find the divergence of the vector field. \overrightarrow{A}=6x^{2}\widehat{a_{x}}+3xy^{2}\widehat{a_{y}}+xyz^{3}\widehat{a_{z}}  at a point P(1,3,6). [04 M]

Ans. To find – Divergence of vector, \overrightarrow{\bigtriangledown }.\overrightarrow{A }

\overrightarrow{\bigtriangledown }.\overrightarrow{A }= \frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}

A_x=6x², A_y=3x², A_z=xyz³

\frac{\partial A_{x}}{\partial x}=12 x

\frac{\partial A_{y}}{\partial y} = 6xy

\frac{\partial A_{z}}{\partial z}=3xyz²

At P(1,3,6) \frac{\partial A_{x}}{\partial x} = 12×1=12

\frac{\partial A_{y}}{\partial y} = 6xy=6×1×3=18

\frac{\partial A_{z}}{\partial z}=3xyz²=3×1×3×6²=324

\overrightarrow{\bigtriangledown }.\overrightarrow{A }=12+18+324= 354

\overrightarrow{\bigtriangledown }.\overrightarrow{A }=354

OR

Q.6. a) Derive the expression for displacement current. Mention four Maxwell’s equations in differential form for time varying fields. [08 M]

Ans. While applying Ampere’s circuital law to a capacitor, Maxwell noticed an inconsistency . According to Ampere’s circuital law,

\oint \overrightarrow{B}.\overrightarrow{dl}=\mu _{o}i

where, i is the current passing through the surface enclosed by a closed loop.

ii) Maxwell made an assumptions that the time varying electric field in the region between the plates is equivalent to a current known as displacement current. If the displacement current id flows in region between plates the problem which arises can be solved.

iii) As the time varying electric field is equivalent to current, it gives rise to magnetic field so we conclude that time varying electric field gives rise to magnetic field.

The modified Ampere’s circuital law includes displacement current as

\oint \overrightarrow{B}.\overrightarrow{dl}=\mu _{0}i_{c}+\mu _{0}i_{d}

If q= Charge on capacitor,

q=Cv=\frac{\varepsilon A}{d}v

= \frac{v}{d}=E is the Electric field

q=εAE

εE=D

q=AD

The current density is given by,

J=\frac{1}{A}\frac{dq}{dt}

J=dD/dt ——————————(1)

b) Derive an expression for numerical aperture in an optical fiber and strain the conditions for propagation. [08 M]

c) Find the attenuation in an optical fiber of length 500 m when a light signal of power 100 mW emerges out of the fiber with a power 90 mW. [04 M]

Module-4

Q.7.a) State and explain Heisenberg’s Uncertainty principle. Show that the electron cannot exist inside the nucleus. [08 M]

Ans. i) Heisenberg’s uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object. This principle is based on the wave-particle duality of matter. Although Heisenberg’s uncertainty principle can be ignored in the macroscopic world (the uncertainties in the position and velocity of objects with relatively large masses are negligible), it holds significant value in the quantum world. Since atoms and subatomic particles have very small masses, any increase in the accuracy of their positions will be accompanied by an increase in the uncertainty associated with their velocities.

ii) In the field of quantum mechanics, Heisenberg’s uncertainty principle is a fundamental theory that explains why it is impossible to measure more than one quantum variables simultaneously. Another implication of the uncertainty principle is that it is impossible to accurately measure the energy of a system in some finite amount of time. The uncertainty is not due to the inadequacy in measuring instruments but it it is already inherent in wave nature.

iii) When we consider the position of particle as a wave group then its position may become uncertain. The particle can be located anywhere in the wave group but its exact location is not known. As narrow as the wave group smaller is the uncertainty in it’s position. In a wave group there is broader range of frequencies and for narrow pulse very large no. of frequencies are present.

iv) We can say that if the wave group is narrower, it will consists of larger number of frequencies and hence wavelength due to which the wavelength becomes more uncertain.

v) As, \lambda =\frac{h}{p} , the uncertainty in wavelength leads to uncertainty in momentum. If the wave group is broader, then there is smaller uncertainty in wavelength and momentum, whereas for narrower wave groups there is large uncertainty in wavelength and momentum. We conclude that, if wave group is narrower then there is small uncertainty in position but large uncertainty in momentum whereas if the wave group is broader then there is large uncertainty in position and small uncertainty in momentum.

vi) The product of uncertainties in the position and momentum is always greater than or equal to h/4π.

The formula is given by, \Delta _{x}\Delta p_{x}\geq \frac{h}{4\pi }. We can write uncertainty principle for the pairs of variables energy-time and angular displacement- angular momentum as \Delta {E}\Delta t\geq \frac{h}{4\pi }  and \Delta {\theta }\Delta L\geq \frac{h}{4\pi }

vii) If the electron is confined to the nucleus having radius of the order of 10^-14m, then the maximum uncertainty in position of the electron will be of order of the radius. Therefore, [{\Delta x}_{max}]=10^{-14}

[{\Delta x}_{max}][{\Delta p}_{min}]=\frac{h}{4\pi } (\Delta p)_{min}=\frac{h}{4\pi (\Delta x)_{max}} (\Delta p)_{min}=\frac{6.63\times 10^{-34}}{4\pi \times 10^{-14}}

(Δp)_min=5.276 × 10^-21 kg.m/s

viii) The moment of electron has to be at least comparable in magnitude to this uncertainty.

p_min≅ ((Δp)_min=5.276 × 10^-21 kg.m/s

From the theory of relativity, the equation of energy is  E=\sqrt{p^{2}c^{2}+m_{o}^{2}c^{4}}

Here, mo²c⁴ << p²c²

E=pc

E_min=5.276 × 10^-21×3×10⁸

= 1.583 × 10^-12 J

E_min=9.9 MeV

ix) Thus, for the electron to exist in the nucleus, its energy must be at least 9.9 MeV. Experimentally it has been observed that when their is beta decay the beta particles never exceed energy about 4 MeV. As the minimum energy of an electron in a nucleus is 9.9 MeV, the electron can never exist in a nucleus.

In this way, it is proved that the electron can never exist in the nucleus.

b) Define spontaneous emission and stimulated emission. Explain the construction and working semiconductor laser. [08 M]

Ans.

• Spontaneous Emission- Spontaneous emission is the process in which a quantum mechanical system (such as a molecule, an atom or a subatomic particle) transits from an excited energy state (E₂) to a lower energy state (E₁)(e.g., its ground state) and emits a quantized amount of energy in the form of a photon(E₂-E₁).

When a large group of atoms emit radiation spontaneously, the emitted radiation contains photons which have their randomly distributed phases, directions and polarization states, even though they have same energy and wavelength.

• Stimulated emission-When the photon having energy E₂-E₁ interacts with the atom in the energy state E₂, the photon forces the atom to undergo transition from the energy state to ground state E₁ giving rise to another photon of same energy E₂-E₁ . This emitted photon will have same phase wavelength, direction and polarization as that of the stimulating photon. Hence the light produced by stimulated emission is coherent. Stimulated emission of radiation is possible from atoms in metastable sates. Spontaneous emission of radiation takes place from excited states with a very small life times.

Semiconductor laser :-A semiconductor laser (LD) is a device that causes laser oscillation by flowing an electric current to semiconductor. The mechanism of light emission is the same as a light-emitting diode (LED). In semiconductor diode, light is generated by the flow of forward current to a p-n junction diode.

Principle: i) When the p-n Junction is forward biased, the electrons from conduction band from N-type gets combined with the valence band holes in the depletion region and then the radiation is emitted.

ii) If the junction is made up of silicon and germanium then the radiation is in infrared region whereas for materials like Gallium Arsenide (GaAs) and Gallium Arsenic Phosphide (GeAsP) the radiation is in the visible region.

iii) The P-N junction is heavily doped and the large amount of current is made to flow through the junction to create population inversion.

iv) For such a junction, by using resonant cavity, we can obtain laser.

Construction- i) In a heavily doped P-n junction diode, the metallic contacts are provided to the P and N types.

ii) The two opposite faces are perpendicular to the plane of the junction and are polished and made parallel to each other. These faces contains the resonant cavity and the laser is obtained through these faces as shown in fig.

iii) The remaining two faces are roughened to prevent the lasing action in that direction.

Working- i) The P and N types are heavily doped, the Fermi level (E_F) in N types lies in the conduction band and in P-type it lies in the valence band. The Fermi level is uniform throughout the unbiased diode as shown in fig.

ii) When the junction is forward biased, the energy level shifts. The width of depletion region decreases due to injection of electrons and holes.

iii) If there is low forward currents, there is recombination of electron-hole occurs and causes spontaneous emission of radiation and the diode acts as a LED.

iv) When the current increases and reaches a threshold value, the population inversion is achieved in the depletion region due to large concentration of electrons in conduction band and holes (i.e valencies) in valence band.

v) The lasing action takes place in the narrow region where population inversion is achieved and it becomes the active region.

vi) This population inversion is produced due to the pumping mechanism which is caused due to the forward bias applied to the junction.

vii) The photons travelling in these junctions along the resonant cavity stimulates the recombination of electron-hole pairs due to which the intensity of coherent light builds up along the axis of the cavity.

viii) The semiconductor lasers have low power consumption and are compact in size. They are very efficient. But it has some disadvantages too like the laser output is less monochromatic and more divergent compared to other lasers.

c) A particle of mass 0.5 meV/C² has kinetic energy 100 eV. Find its de Broglie wavelength, where C is the velocity of light. [04 M]

Ans.- Given – m= 0.5, E=100 eV,

Formula- \lambda =\frac{h}{\sqrt{2mE}}

Solution- By de Broglie hypothesis,

\lambda =\frac{h}{\sqrt{2mE}}

h=6.63 × 10^-34 J.S, m=0.5 MeV/C²= \frac{0.5\times 1.6\times 10^{-13}}{(3\times 10^{8})^{2}}=8.9 \times 10^{-31} kg

E=100 V= 100 × 1.6 × 10^-19 J= 1.6 × 10 ^-17 J  = 1.6 × 10 ^-17 J

\lambda =\frac{6.63\times 10^{-34}}{\sqrt{2\times 8.9\times 10^{-31}\times 1.6\times 10^{-17}}=1.24 \times 10^{-10}m

λ=1.24 Å

OR

Q.8. a) Assuming the time independent Schrodinger wave equation, discuss the solution for a particle in one dimensional potential well of infinite height. Hence obtain the normalized wave function. [08 M]

Ans. i) By knowing the potential energy of the particle to be infinite at the end beyond the walls, the physical problem of particle confined between rigid walls can be converted into a problem of potential distribution. AS shown in fig.

V=∞ for x≤0 and for x≥L

ii) As the potential energy of the particle is constant so we can take it as zero for our convenience. i.e. V=0 for 0<x<L

iii) As the particle is inside the infinite potential well, the particle does not exist at the walls and beyond them.

ψ=0 for x≤0 and x≥L. The wave function ψ exists only when 0<x<L .

Schrodinger’s time independent equation is given by

\bigtriangledown ^{2}\Psi +\frac{8\pi ^{2}m}{h^{2}}(E-V)\Psi =0———-(1)

iv) It is the problem in one dimension along x-axis, ∇²ψ can be replaced by  \frac{\partial ^{2}\Psi }{\partial x^{2}} which can also be replaced as \frac{\d ^{2}\Psi }{\d x^{2}} as ψ is a function of x only. Also, by substituting V=0 for 0<x<L in eq. 1,

\frac{d^{2}\Psi }{dx^{2}}+\frac{8\pi ^{2}m}{h^{2}}E\Psi =0

Let, k=\frac{8\pi ^{2}m}{h^{2}}E—————–(2)

\frac{d^{2}\Psi }{dx^{2}}+K^{2}\Psi =0———–(3)

The general solution of equation (4) is

Ψ=A sin Kx+bcos Kx———-(4)

where A and B are arbitrary constant which are to be determined using boundary conditions.

v) The first boundary condition is given by Ψ=0 at x=0

From equation (4), we get

0=A sin 0 +B cos 0

B=0

From equation (4) we get, Ψ=A sin Kx———-(5)

vi) The second boundary condition is given by, Ψ=0 and x=L, we get from eq. (5) 0=A sin KL

A≠0 as for A=0, Ψ=0 for all values of x which will mean that particle does not exist inside the box.

Sin KL=0, KL=nπ, n=1,2,3…

n≠0 as n=0 ⇒Ψ=0 for all nvalues of x which is not possible

K=nπ/L —-(6)

From eq. (2) and (5) we get,

\frac{8\pi ^{2}mE}{h^{2}}=\frac{n^{2}\pi ^{2}}{L^{2}}

As energy ‘E’ depends on ‘n’ we use n as suffix for ‘E’

E_{n}=\frac{n^{2}h ^{2}}{8mL^{2}}————(7)

where, n=1,2,3,..

From the above equation, the smallest value of energy that the particle can have is

E_{1}=\frac{h ^{2}}{8mL^{2}}

It is non-zero. This all contradicts the classical mechanics, according to which we can say that the particle can have zero energy.

vii) The other possible values can be E_{2}=\frac{4h ^{2}}{8mL^{2}}, E_{3}=\frac{9h ^{2}}{8mL^{2}}, etc.

These energy values are discrete. They are not continuous  as we had expected from classical mechanics. Thus, we can say that the according to the quantum mechanics, the particle inside a rigid box cannot have all values of energy, but the discrete values which are given by eq. (20

These discrete energy values are known as energy eigen values. These energy levels are shown in fig.

Now obtaining normalized wave function,

viii) Substituting eq. (60 in eq. (5) we get,

\Psi =A sin \left ( \frac{n\pi x}{L} \right )———(8)

The complex conjugate of ψ is

\Psi^{*} =A sin \left ( \frac{n\pi x}{L} \right )

To normalize wave function e have to find \int_{0}^{L}\Psi \Psi ^{*}dx

\int_{0}^{L}\Psi \Psi ^{*}dx=\int_{0}^{L}A^{2}sin^{2}\left ( \frac{n\pi x}{L} \right )dx =A^{2}\int_{0}^{L}\left ( \frac{1-cos\frac{n\pi x}{L}}{2}\right )dx A^{2}\int_{0}^{L}\left [ \frac{1-cos(\frac{n\pi x}{L})}{2} \right ].dx \frac{A^{2}}{2}\left [x- \frac{sin(\frac{n\pi x}{L})}{\frac{n\pi }{L}} \right ]^{L}_{0} =\frac{A^{2}}{2}[L-0] \int_{0}^{L}\Psi \Psi ^{*}.dx=\frac{A^{2}L}{2}

ix) Considering R.H.S as N² i.e

N^{2}=\frac{A^{2}L}{2}

or N=A\sqrt{\frac{L}{2}}

Now, the normalized wave function ψ_n can be obtained as,

\Psi _{n}=\frac{\Psi }{N} \Psi _{n}=\frac{A sin (\frac{n\pi x}{L})}{A\sqrt{\frac{L}{2}}} \Psi _{n}=\sqrt{\frac{2}{L}}sin \left ( \frac{n\pi x}{L} \right )

x) The normalized 3wave function are called eigen functions.

\Psi _{n}^{*}=\sqrt{\frac{2}{L}}sin \left ( \frac{n\pi x}{L} \right )

The probability function is given by,

P(x)=|Ψ_n|²=Ψ_nΨ_n^*= \frac{2}{L}sin^{2}\left ( \frac{n\pi x}{L} \right )————————(10)

The fig below shows the wave function and probability function for n=1,2,3

b) Derive the expression for energy density interns Einstein’s co-efficient. [08 M]

Ans.i)  consider the system of atoms which has ground state E₁ and excited state energy E₂ and having no. of densities of atoms as N1 and N2 respectively. When the photons of frequency v, will be incident on the system then there will be induced absorption. The photon frequency is given by, v=\frac{E_{2}-E_{1}}{h}

ii) The  Photons absorption rate will be directly proportional to number density N₁ of the atoms in ground state and the energy density E_v is in the frequency range v to v+dv in incident radiation.

Rate of absorption ∝ N₁ E_v

Rate of absorption = B₁₂ N₁ E_v

Here, B₁₂ is the constant and this constant is known as the Einstein’s coefficient of induced absorption.

iii) The atoms in excited state can come down to the ground state either by spontaneous emission or through stimulated emission of radiation.

In the spontaneous emission the rate of transition of atoms from the excited state E₂ to Ground state E₁ does not depend on the energy density in the incident radiation and is proportional only to the number density of atoms in thee excited state.  i.e.
Rate of spontaneous emission ∝ N₂

Therefore, Rate of spontaneous emission = A₂₁ N₂, Here A₂₁ is constant and it is named as Einstein’s coefficient of spontaneous emission.

In case of stimulated emission a photon of frequency, v=\frac{E_{2}-E_{1}}{h} is required to stimulate the atoms.

iv) Hence, the rate of change of stimulated emission is proportional to energy density E_v and the number density N₂ of atoms in the excited energy state E₂ i.e.

Rate of stimulated emission ∝ N₂Ev

Rate of stimulated emission = B₂₁ N₂ E_v, where B₂₁ is constant and  is known as Einstein’s coefficient of stimulated emission.

In Thermal equilibrium state the rate of transition of atom from E₁ to E₂ must be equal to total rate of transition from E₂ to E₁.

The Rate of absorption = Rate of spontaneous emission + Rate of stimulated emission

B₁₂ N₁ E_v=A₂₁ N₂+B₂₁ N₂ E_v

Dividing by N_1, we get

B_{12}E_{v}=A_{21}\frac{N_{2}}{N_{1}}+B_{21}\frac{N_{2}}{N_{1}}E_{v} E_{v}{B_{12}-B_{21}\frac{N_{2}}{N_{1}}}=A_{21}\frac{N_{2}}{N_{1}} E_{v}=\frac{A_{21}\frac{N_{2}}{N_{1}}}{{B_{12}-B_{21}\frac{N_{2}}{N_{1}}}} E_{v}=\frac{A_{21}}{{B_{12}\frac{N_{1}}{N_{2}}-B_{21}}} E_{v}=\frac{A_{21}}{B_{21}}\left [ \frac{1}{(\frac{B_{12}}{B_{21}})(\frac{N_{1}}{N_{2}})-1} \right ]

v) From Maxwell- Boltzman distribution,

\frac{N_{2}}{N_{1}}=e^{-(E_{2}-E_{1})/kT}=e^{-hv/kT} \frac{N_{1}}{N_{2}}=e^{-hv/kT}

E_{v}=\frac{A_{21}}{B_{21}}\left [ \frac{1}{(\frac{B_{12}}{B_{21}})(e^{hv/kT})-1} \right ] —————————-(1)

vi) Comparing this equation to the energy density from planck’s law,

E_{v}=\frac{8\pi hv^{3}}{c^{3}}\left [ \frac{1}{(e^{hv/kT})-1}}\right ] ——————————-(2)

We get, \frac{B_{12}}{B_{21}}=1

B₁₂=B₂₁ ———(3)

\frac{A_{21}}{B_{21}}=\frac{8\pi hv^{3}}{c^{3}}————(4)

Eq. 3 indicates that the probability of induced absorption is same as the probability of stimulated emission.

From eq. 4,

\frac{A_{21}}{B_{21}}\propto v^{3}

For large v, A₂₁>> B₂₁. As v=\frac{E_{2}-E_{1}}{h} for large energy difference between the ground state and excited state, the probability of spontaneous emission is much larger than the probability of stimulated emission.

vii) To build the lasers in the ultraviolet or x-ray region would be much more difficult than to build lasers in visible or infrared regions. The process of stimulated emission becomes significant at lower frequencies.

viii) Consider a system of atoms with an excited state and having a large life span so the spontaneous emission can be neglected. As rate of induced absorption is proportional to N₁ and rate of stimulated emission is proportional to N2, fr N1>N2, the dominant vprocess would be induced absorption.

ix) This may decrease the number of photons with time. For making the process of stimulated emission dominate over the process of induced  absorption, the no. of photons increases with time, the no. density of atoms in excited state has to exceed he no. density in the ground state.

This state of system is known as Population inversion.

c) The ratio of population of two energy levels is 1.059× 10 -30 . Find the wavelength of light emitted by spontaneous emissions at 330 K. [04 M]

Ans. Given- N₂/N₁=1.059×10^-30, T=300 K

Solution- \frac{N_{2}}{N_{1}}=e^{-(E_{2}-E_{1})/kT}

\frac{N_{2}}{N_{1}}=1.059\times 10^{-30}

K=1.38 × 10^-23 JK^-1, T=330 K

1.059\times 10^{-30}=e^{-(E_{2}-E_{1})/(1.38\times 10^{-23}\times 330)} 9.515\times 10^{-29}=e^{-(E_{2}-E_{1})/(4.554\times 10^{-21}) ln 9.515\times 10^{-29}=\frac{E_{2}-E_{1}}{4.554\times 10^{-21}}

E₂-E₁=3.14× 10-19 J

Also, E_{2}-E_{1}=\frac{hc}{\lambda }=\frac{6.61\times 10^{-34}\times 3\times 10^{8}}{\lambda }

E_{2}-E_{1}=\frac{1.989\times 10^{-25}}{\lambda} J

3.14 \times 10^{-19}=\frac{1.989\times 10^{-25}}{\lambda}

λ=6.334×10^-7 m

λ=633.4 nm

Module 5

Q.9. a) Give the assumptions of quantum free electron theory. Discuss two success of quantum free electron theory. [08 M]

Ans. The assumptions of the quantum free electron theory are as follows:-

• Inside the metal, the valence electrons is free to move.
• As, the potential is constant inside the metal, the electrons are confined to the metal by potential barrier at the boundaries.
• There is negligible electrostatic force of attraction between the free electrons and the ion cores.
• According to the pauli’s exclusion principle, the distribution of electrons are quantized and the distribution of electrons are in the allowed discrete energy levels. It prohibits more than one electrons in a single quantum state.

The success of quantum free electron theory are as follows:

i) The quantum free electron theory provides the correct dependence of resistivity on temperature.

ii) This theory gives accurate molar specific heat at constant volume for the free electrons in a metal which is low as compared to that predicted by classical free electron theory.

b) What are polar and non-polar dielectrics? Explain types of polarisation. [08 M]

Ans. Polar dielectrics are materials which have an inherent dipole moment. In polar dielectrics, the positive and negative charges are separated by a small distance in the absence of electric field.

Non-Polar Dielectrics: Non-Polar dielectrics are materials which do not have an inherent dipole moment.

i) When the non-polar dielectric materials is given an external electric field then the charges present in the dielectric materials gets separated. It means the negative charge centre of all the electrons in the atom or molecule gets separated from the positive charge center of the nucleus, thus creating dipole moment. This induced dipole moment in non-polar dielectrics is called as polarization.

ii) In polar dielectrics polarization refers to partial alignment of existing dipoles on application of external electric field.

iii) In ionic solids like Nacl and KCL, polarization occurs as a result of the displacement between positive and negative ions in an external field.

iv) Due to the dipoles included, an electric field is produced in the dielectric. This internal electric field is directed opposite to the external electric field. The effect of polarization is to weaken the externally applied electric field.

v) In non-polar dielectrics there is absence of external field a shown in the fig.

vi) In the presence of externally applied field E_o, dipoles are induced, due to the displacement of positive charge centres in the direction of E_o and negative charge centres opposite to E_o.

vii)  Due to this polarization, the surface charges are produced on the dielectric slab. Positive charge develops on the rights ide of the slab surface and negative charges are produced on the left surface.

viii) The electric field Ep develops opposite to E_o due to these surface charges. Hence the net electric field E will be less than E_o. Thus the presence of dielectric reduces the electric field.

ix) In polar dielectric, as electric dipoles are randomly oriented, there is no electric field in the dielectric.

x) Due to thermal vibration the complete alignment is prevented . Surface charges are produced due to alignment of dipoles giving rise to electric field E_p in opposite direction to E_o. The net electric field E is reduced like in non-polar dielectrics.

Types of polarisation- There are four types of polarization.

i) Electronic or induced polarization– When an atom or molecule is subjected to the electric field, the positive charge centers of the nucleus shifts in the direction of the electric field whereas negative charges centre corresponding to the electrons which is in the direction opposite to electric field.

In this way, dipoles in the dielectric is created when it is placed in electric field. This process of dipole formation is known as electronic or induced polarization.

ii) orientation polarization– i) polar dielectric contains molecules having dipole moment even in absence of any applied electric field. These dipoles are randomly oriented.

ii) When there is presence of an external electric field, the dipoles get partially aligned along the field resulting in the development of an electric field opposing the external field. This type of polarization is known as orientation polarization.

iii) Ionic or atomic polarization– i) If ionic solids are placed in electric field, there is displacement of positive ions in the direction of electric field and negative ions opposite to electric field which results in the formation of dipoles.

ii) This type of polarization is known as ionic or atomic polarization.

iv) Space charge polarization or Interfacial polarization- Due to large changes in resistivity, the charges gets accumulated at interfacial boundaries. This creates dipole moments. this type of polarization is small compared to other three and hence is neglected in most cases.

c) Calculate the probability of an electron occupying an energy level 0.02 ev above the Fermi level at 200 K and 400 K in a material. [04 M]

Ans. Given – E-E_F=0.02 eV,    T=200 K,  T=400 K

Solution- The fermi distribution function is given by

f(E)=\frac{1}{1+e^{E-E_{F}/KT}}

E-E_F=0.02 eV

k=1.38×10^-23 J/K = 8.625 × 10^-5 eV/K

For, T=200 K,

f(E)=\frac{1}{1+e^{\frac{0.02}{8.625\times 10^{-5}\times 200}}}

f(E)=0.239

For T=400 K

mf(E)=\frac{1}{1+e\frac{0.02}{8.625\times 10^{-5}\times 400}}

f(E)=0.359

Note- the probability of occupancy of energy levels above fermi level increases on increasing the temperature.

OR

Q.10. a) Define internal field. Mention the expressions for internal field, for one dimension, for three dimensional and Lorentz field for dialectics. Derive Clasusius- Mossotti equation. [08 M]

Ans. i) In dielectric solids, the atoms or molecules experience not only the external applied electric field but also the electric field produced by the dipoles. The resultant electric field acting on the atoms or molecules of dielectric substance is called the local field or an internal field.

ii) Lets take example, Consider a linear array of equidistant atoms in an electric field  E as shown in fig. The externally applied electric field E polarizes the atoms with dipole moments.

iii) WE can see that positive charge centres and displaced towards right ( in the direction of applied field) and negative charge centres to the left  due to which the dipole moment vector is directed from left to right.

iv) The electric fields at location X due to dipoles at A₁ and A₂ will have same magnitude and they both will be directed towards right. Hence, the net electric field at X due to both A₁ and A₂ will be,

E_{1}=2\left [ \frac{\mu }{2\pi \varepsilon _{0}d^{3}} \right ] E_{1}=\frac{\mu }{\pi \varepsilon _{0}d^{3}}

v) The electric field at X due to dipoles at B₁ and B₂ will be,

E_{2}=2\left [ \frac{\mu }{2\pi \varepsilon _{0}(2d^{3})} \right ] E_{2}=\frac{\mu }{\pi \varepsilon _{0}(2d^{3})}

Similarly, E_{3}=\frac{\mu }{\pi \varepsilon _{0}(3d^{3})}

E_{4}=\frac{\mu }{\pi \varepsilon _{0}(4d^{3})}

vi) The total electric field due to induced dipoles at X will be,

E_p=E₁+E₂+E₃+——

E_{p}=\frac{\mu }{\pi \varepsilon _{0}(d^{3})}+\frac{\mu }{\pi \varepsilon _{0}(2d^{3})}+\frac{\mu }{\pi \varepsilon _{0}(3d^{3})}+------ E_{p}=\frac{\mu }{\pi \varepsilon _{0}(d^{3})}\left [ 1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+---- \right ] E_{p}=\frac{\mu }{\pi \varepsilon _{0}(d^{3})}\sum_{n=1}^{\infty }\frac{1}{n^{3}} \sum_{n=1}^{\infty }\frac{1}{n^{3}}=1.2 E_{P}=\frac{1.2\mu }{\pi \varepsilon _{0}d^{3}}

vii) The internal field Ei at X is the resultant of applied electric field E and the net electric field E_p due to all the induced dipoles i.e.

E_i=E+E_p

E_{i}=E+\frac{1.2\mu }{\pi \varepsilon _{0}d^{3}}———-(1)

viii) We know that,

μ=α_e E_i

Where, α_e is the electronic polarizability.

E_{i}=E+\frac{1.2\alpha _{e}E_{i} }{\pi \varepsilon _{0}d^{3}}

E_{i}=\frac{E}{1-\frac{1.2\alpha _{e}E_{i} }{\pi \varepsilon _{0}d^{3}}}———-(2)

viii) The equation above indicates the internal field E_i of a linear array of atom interms of the electronic polarization α_e and the interatomic distance d.

ix) The internal field is greater than the applied field E as the denominator in the above equation is less than 1. The internal field increases with increase in electronic polarizability α_e and decrease in interatomic distance d. The internal field in three dimensions depends upon the crystal structure and is given by,

E_{i}=E+\frac{\gamma P}{\varepsilon _{0}}

where γ is constant known as internal field constant and depends on the crystal structure.

For cubic, crystal structure, γ=1/3

E_{i}=E+\frac{P}{3\varepsilon _{0}}———————–(3)

x) This internal field in cubic crystal structures is known as Lorentz field. From the above equation we come to the point that internal field increases with increase in polarization P, i.e. the induced dipole moment per unit volume.

xi) The Clausius- Mososotti equation is valid for elemental solids like diamonds, sulphur, germanium etc. they do not have ions or permanent dipole moments and have a cubic structure for which the Lorentz field is applicable.

These solids nhave only electronic polarization.

P=Nα_eE_i———————(4)

The Lorentz field E_i is given by,

E_{i}=E+\frac{P}{3\varepsilon _{0}}

Putting values of E_i in Eq. 4,

P=N\alpha _{e}E+\frac{N\alpha _{e}P}{3\varepsilon _{0}}

As, P=ε₀(ε_r-1)E

\varepsilon _{0}(\varepsilon _{r}-1)E=N\alpha _{e}E+\frac{N\alpha _{e}}{3\varepsilon _{0}}\times \varepsilon _{0}(\varepsilon _{r}-1)E \varepsilon _{0}(\varepsilon _{r}-1)=N\alpha _{e}E+\frac{N\alpha _{e}}3 (\varepsilon _{r}-1) \varepsilon _{0}(\varepsilon _{r}-1)=N\alpha _{e}\left [ 1+\frac{\varepsilon _{r}-1}{3} \right ] \varepsilon _{0}(\varepsilon _{r}-1)=N\alpha _{e}\left [\frac{\varepsilon _{r}+2}{3} \right ]

\frac{N\alpha _{e}}{3\varepsilon _{0}}=\frac{\varepsilon _{r}-1}{\varepsilon _{r}+2}——-(5)

The above equation is known as Claussius-Mossotti equation which is valid only for elemental solids having cubic structures.

b) Describe Fermi level in an intrinsic semi conductor and hence obtain the expression for fermi energy in terms of energy gap of intrinsic semiconductor. [08 M]

Ans. i) The probability of occupation of energy levels in valence band and conduction band is called Fermi level. At absolute zero temperature intrinsic semiconductor acts as perfect insulator. However as the temperature increases free electrons and holes gets generated.

In intrinsic or pure semiconductor, the number of holes in valence band is equal to the number of electrons in the conduction band. Hence, the probability of occupation of energy levels in conduction band and valence band are equal. Therefore, the Fermi level for the intrinsic semiconductor lies in the middle of forbidden band.

ii) Fermi-Dirac probability function is given by

P(E)=\frac{1}{1+e^{\frac{E-E_{F}}{kT}}}

This function gives the probability that an electron has energy ‘E’ .

ii) It is assumed that width of valence band and conduction band is very small compared to the forbidden energy gap E_g. As their width is small they are represented by single values of energy E_v and E_c respectively.

iii) The probability that an electron is in valence band is the probability that it has energy gap E_v and is represented by P(E_v) which is given by

P(E_{v})=\frac{1}{1+e^{\frac{E_{v}-E_{F}}{kT}}}

iv) Similarly, the probability that the electron is in conduction band is given by,

P(E_{c})=\frac{1}{1+e^{\frac{E_{c}-E_{F}}{kT}}}

If at any temperature ‘T’

n_c= Number of electrons in conduction band

n_v=Number of holes in valence band

N= Total no. of electrons in valence and conduction band

N=n_v+n_c

\frac{n_{v}}{N}+\frac{n_{c}}{N}=1

As, P(E_{c})=\frac{n_{c}}{N}

and, P(E_{v})=\frac{n_{v}}{N}

P(E_v)+P(E_c)=1

Putting values from equations (2)  and (3),

\frac{1}{1+e\frac{E_{v}-E_{F}}{kT}}+\frac{1}{1+e\frac{E_{c}-E_{F}}{kT}}=1 {1+e\frac{E_{c}-E_{F}}{kT}+{1+e\frac{E_{v}-E_{F}}{kT}=[{1+e\frac{E_{c}-E_{F}}{kT}][{1+e\frac{E_{v}-E_{F}}{kT}]

2+e^(E_c-E_F)/kT+e^(E_v-E_F)/kT=1+e^(E_v-E_F)/kT+e^(E_c-E_F)/kT+e^{(E_v+E_c-2E_F)/kT}

1=e^{(E_v+E_c-2E_F)/kT}

(E_v+E_c-2E_F)/kT=0

(E_v+E_c-2E_F)=0

E_{F}=\frac{E_{v}+E_{c}}{2}

Thus, the Fermi level lies exactly at the centre of the forbidden gap as shown in fig.

c) An elemental solid dielectric material has polarizability 7 × 10 ^-40 Fm² . Assuming the internal field to be Lorentz field, calculate the dielectric constant for the material if the material has 3 × 10²⁸ atoms/m³. [04 M]

Ans. By Clausisus-Mossotti equation,

\frac{N\alpha _{e}}{3\varepsilon _{0}}=\frac{\varepsilon _{r}-1}{\varepsilon _{r}+2}

N=3×10²⁸ atoms/m³

α_e=7×10^-40 F-m²

ε₀=8.85×10^-12 F/m

\frac{\varepsilon _{r}-1}{\varepsilon _{r}+2}=\frac{3\times 10^{28}\times 7\times 10^{-40}}{3\times 8.85\times 10^{-12}}

\frac{\varepsilon _{r}-1}{\varepsilon _{r}+2}=0.79

\varepsilon _{r}-1=0.79 \varepsilon _{r}+1.58

ε_r=12.286

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μm