B.

Tech 1st Sem Examination, 2013

PHYSICS-1 ( BS- 1102 ); Question Code: C-611

Full Marks – 70 Time- 3hrs

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1. Answer the following questions: ( 2 x 10 = 20 )

(a) Distinguish between velocity resonance and amplitude resonance.

Soln:

 Velocity resonance occurs when the velocity of simple harmonic oscillator is maximum at its
equilibrium position.

NIST
Amplitude resonance occurs when the amplitude of the oscillator is maximum. The
amplitude becomes maximum when the frequency of the external force is equal to the natural
frequency of the vibrating body.

 velocity resonance :There is no phase difference between applied force & velocity.

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amplitude resonance :Maximum amplitude lags behind the applied force by π/2 radian.

(b) Mention the conditions for the formation of stationary waves.

Soln:

Solutions
Stationary waves are produced when two waves of same frequency, wavelength and amplitude
travelling with same speed in opposite directions in a region superpose each other.

(c) The amplitudes of light waves emerging from two slits in Young’s experiment is in the ratio1:2.
Find the intensity ratio of the interference pattern.

Soln:

=1 =2

Maximum intensity =( + )2 = ( 1 + 2 ) 2 = 9

Minimum intensity =( - )2 = ( 1 - 2 ) 2 = 1

: = 9:1

( d ) Why diffraction pattern is not generally observed with an extended source of light?

Soln: A wide slit is equivalent to large number of narrow slits. If a wide slit is used, each narrow slit
will produce its own diffraction pattern. Due to overlapping of diffraction patterns on the screen,
general illumination is observed. Hence the diffraction pattern cannot be observed.
(e) What are the essential conditions for Fraunhofer diffraction?

Soln:

The source and the screen are effectively at infinite distances from the aperture causing diffraction.
One convex lens is used to make the incident rays parallel and another convex lens is used to focus
the diffracted rays. The width of the aperture must be comparable with the wavelength of the
incident light.

( f ) When a beam of light is polarised, does its intensity vary?

Soln:

yes, the intensity do vary.

( g ) Distinguish between conduction and displacement current.

Soln:

NIST
(1) Conduction current is real current where as displacement current is a fictitious current.
(2) The conduction current is due to flow of electrons in a conducting medium where as
displacement current is due to time varying electric field.
(3) The conduction current obeys ohm’s law. That is, it depends on the resistance and potential

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difference of the conductor. The displacement current depends on the permittivity of the
medium and the rate at which the electric field changes with time.

(h) Write the differential and integral form of Faraday’s law.

Solutions
Soln: The differential form of Faraday’s law of electromagnetic induction is

⃗
⃗⃗⃗ ⃗ + =0
The integral form of Faraday’s law is


 E  dl    BdS
c t S

⃗ = Intensity of electric field

⃗ = Magnetic induction

S = surface enclosed by the loop, dl = small length of the loop C

ds = infinitesimal surface area

( i) What is the phase difference between ⃗ and ⃗ ? Explain .

Soln: The phase difference between ⃗ and ⃗ is zero in free space as they attain maximum
and minimum value at a time.
 E/B =C ; C=velocity of light in vacuum

The ratio of E/B is constant showing that the vectors E and B are always in phase with each other.

In case of conducting medium there is some phase difference between ⃗ and ⃗ due to the
dissipative term in the wave equation. The dissipation of magnetic field and electric field are
different in a medium.

(j) What is the physical significance of threshold frequency?

In case of photoelectric effect, the emission of photocurrent starts only when the frequency of
incident radiation is greater than the threshold frequency.

2.(a) Derive the general differential equation of the wave motion. (7)

The equation of a wave travelling along x-direction is given by

NIST
( x, t ) = A Sin ( kx –ωt + ) ................................................................................................ (1)

( x, t ) = wave function at the position ‘x’ at time ‘t’.

⃗ = wave vector : ω = Angular frequency ; A = Amplitude of the wave

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From eq(1)

= - ωA cos ( kx –ωt + ) ; = - ω 2 A Sin ( kx –ωt + ) = - ω 2 Ψ ..............................(2)

= Ak cos ( kx –ωt + ) ; = - A k2 Sin ( kx –ωt + ) = - k2 Ψ .....................................(3)

Solutions
Using the expression of Ψ from eq(3) in eq(2) we have,

= ( ω 2 / k2) -

=> = v2 - ( as ω/k = v ) .................................................(4)

Where ‘v’ is the velocity of the wave.

Eq(4) is called the differential equation of the wave motion along x-axis. In general Eq(4) can be
written as

= v2 Ψ .................................................................(5)

(b) Mention the different types of progressive waves. ( 3)

Soln:

Progressive waves are of three types:-

(i) Longitudinal waves

 Ex:- sound waves in solids ,liquids and gases.

(ii) Transverse waves

Ex:-Electromagnetic waves, waves in a stretched string

(ii) Surface waves

Ex:- waves on the surface of water, Rayleigh wave

3(a) Discuss the formation of Newton’s rings by reflected light. (6)

Soln:

A plano-convex lens L of large focal length (or radius of curvature) is placed on a plane
glass plate P with its convex surface touching the plate at O. An air film of increasing thickness is
formed between the convex lens and the glass plate. The thickness of the air film is zero at the point
of contact ‘o’, and increases towards the rim.

NIST
A monochromatic source of light, such as a sodium vapour lamp, is placed
behind a small circular aperture S, made on the opaque plate. S is placed at the focus of convex lens
L1. A parallel beam of horizontal rays emerge from this lens.

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A glass plate G mounted at an angle of 450 to the horizontal partially reflects the rays
incident on it. At an angle of incidence of 450, the reflected rays travel vertically downward and are
incident normally on the plane surface of the convex lens L. Hence the incident rays pass through the
upper surface of L without deviation. At the curved surface of the lens, a part of incident light is
partially reflected back but the other part travels further and is reflected from the plane glass plate

Solutions
P.

Both the vertically reflected beams, which serve as two coherent sources, unite but the
second part has travelled through a longer distance. Therefore, path difference is introduced
between these two beams. When viewed through a travelling microscope, T, interference pattern
localized in the air film is observed.

Since the air film is symmetrical about the point of contact O, the fringes which follow lines
of equal thickness will be concentric alternate bright and dark rings. The centre of the ring system is
dark. These rings are called Newton’s rings.
 NIST
S: Monochromatic source; L1: Convex lens; T: Telescope; G: glass plate;

L: Plano-convex lens; P: plane glass plate

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3 (b) Describe the experimental arrangement and give necessary theory. (2)

Soln:

Diagram of 2(a)

Solutions
The condition for bright ring is 2t = (2n – 1) λ/ 2; where n = 1,2,3........ ..........(1)

The condition for dark ring is 2t = n λ ; where n = 0,1,2,3........ ..........(2)

t = Thickness of the air film ; λ = wavelength of incident light.

In terms of the diameter Dn of the ring, the conditions are

Dn2 = 4nRλ for dark ring ........................................(3)

and Dn2 = 2R (2n- 1) λ for bright ring .......................................(4)

3 (c) Why Newton’s rings are circular? (2)

Soln:

The conditions for bright and dark Newton’s fringes are given by

2t =(2n-1 )λ /2 (bright fringe) Where n=1,2,3.................

2t =n λ(dark fringe) Where n=0,1,..............

These conditions depend on the thickness t of air film. Hence the bright or dark fringes of any
particular order n will occur for constant value of t.The thickness of the air film between plano-
convex lens and plane glass plate accounts for the path difference. Since the thickness of the air film
is constant along a circle, the dark and bright fringes are in circles. That is why the interference
fringes are circular.

4.(a) What is a Zone plate? Derive an expression for its focal length. (5)

Soln:

A zone plate is an optical device that is used to verify the correctness of Fresnel’s method of dividing
a wave front into half period zones .It is a special diffracting screen designed to obstruct the light
from the alternate half period zones. It is a transparent plate on which a series of concentric circles
are drawn with their radii proportional to the square root of natural numbers .The alternate annular
zones are blocked .The zone plate behaves like a convex lens and produces an image of a source of
light on the screen placed at a suitable distance on the other side.

Expression for focal length:-

NIST
In the following fig. AB represents the section of a positive zone plate. That is, odd zones are
transparent and even zones are opaque to light. O is the centre of zone plate.
M1,M2,M3,......represent outer limits of the 1st, 2nd,3rd,........zones respectively. A line passing through
the centre O of the zone plate and perpendicular to it is called the axis of the zone plate .S is a point
source of light on the axis of the zone plate. A point P is taken on the axis of the zone plate on the

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other side. SO=a=object distance, PO=b=image distance.

The point P is chosen in such a way that

SM1+M1P=SO+OP+ λ/2

Solutions
SM2+M2P= SO+OP+2 λ/2

.......................................

SMn+MnP= SO+OP+n λ/2.............(1)

 Let rn=OMn=radius of nth circle on the zone plate

SMn= SO2  OM n 2 = a 2  rn 2 =a(1+rn2/a2)1/2=a+rn2/2a (rn<<a)

MnP= PO 2  OM n 2 = b2  rn 2 =b(1+rn2/b2)1/2=b+rn2/2b(rn<<b)

Substituting these values in equation ( 1 )

( ) ( )= ( )
2 2 2

=>( ) ( )=

=>(1/a)+(1/b)=( / )……………………..(2)

NIST
Comparing eq(2) with eq. 1/u+1/v=1/F of convex lens we get,

1/f= / or f= / ………………………(3)

Where f is the primary focal length of the zone plate.

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Q 4b Show that the radii of its half period zones are proportional to the square root of natural
numbers. (3)

Soln:

Solutions
We have (1/a)+(1/b)=( / )

=> (a+b/ab)=( / )

=> = √( ) √

Since a,b & are constants for a given object, image and light, we have,

√ Where n=1,2,………..

Q4(c) Show that a zone plat has multiple foci

Soln:

From the theory of the zone plate, we have

(1/a)+(1/b)=( / ) = 1/

If a=∞; we have f=b= /

For the point P, even zones are opaque. The resultant amplitude A at p is given by,

A=A1+A3+A5+……….
Where A1,A3,A5,……..are the amplitudes of light waves arriving at P from 1st,3rd,5th etc zones
respectively. Hence, the intensity at P increases.

Let us consider a point P1 closer to the zone plate at a distance b/2=f/2 from the zone plate. Each
transparent zone allows two consecutive half period zones, while the next half period zones are
blocked.

So, the resultant amplitude at P1 is,

A’ =(A1-A2)+(A5-A6)+…….

Since A1 ≈A2, A5 ≈ A6,…….,hence we have A’=0

Hence the point P1 is dark.

Similarly, consider a point P2 further closer to the zone plate at a distance b/3=f/3. The three
consecutive half period zones are allowed through each transparent zone of the zone plate, while

NIST
the next three half period zones are blocked.

The resultant amplitude at P2 is

A’’= (A1-A2+A3)+(A7-A8+A9)+……

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Since A1+A3 ≈ 2A2; A7+A9 ≈ 2A8, ……we have

A’’=A2+A8+…..

That is, A’’ have finite value.

Solutions
Hence the point P2 will be a bright point. Similarly at a distance b/4=f/4 a dark point is obtained,
whereas at a distance b/5=f/5 a bright point observed.

Thus, if light is incident normally on a zone plate, then the corresponding focal points are at
distances f,f/3,f/5,……etc from the zone plate. Hence a zone plate has multiple foci.

5 (a) How planes are polarized, elliptically polarized & circularly polarized light produced
experimentally? (7)

Soln:

: Production of plane polarized light : experimentally plane polarized light can be produced by
following methods

 By reflection: When light falls on a mirror at Brewster’s angle that is at 570, the reflected
light becomes completely plane polarized.
 By refraction: When light falls on a glass slab at polarizing angle, the transmitted light

NIST
becomes partially plane polarized. So allowing this light to pass through a pile of plates will
make the emergent light completely plane polarized.

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
Solutions
By double refraction: When an un polarized light enters into a Nicolprism, the ordinary light
undergoes total internal reflection and gets absorbed at the side surfaces and the extra
ordinary light emerges out as plane polarized light.
 Production of elliptically and circularly polarized light : To produce elliptically polarized light the
un polarized light is passed through a nicol prism. A quater wave plate is placed before the
emergent plane polarized light, in such a way that the short diagonal of the polariser makes
angle θ with the optic axis of the quater wave plate. When the plane polarized light enters into
the quater wave plate, it splits into two polarized beams perpendicular to each other. In this
case if E is the amplitude of plane polarized light before it enters into the quater wave plate,
then the amplitudes of splited beams are Ex=E cos θ , Ey= E sin θ . When θ ≠450 amplitudes of
the splited beams are different. Hence emergent light is elliptically polarized and for θ=450, it
become circularly polarized as the amplitudes are equal.

NIST OR (The following answer can also be written)

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Production of plane polarized light:

Solutions
ABCD represents the principal section of Nicol prism. Let a ray P of unpolarized light parallel to the
longer side AD be incident on the face AB of Nicol prism. On entering the Nicol prism, it is broken up
into two refracted rays the O-ray & the E-ray by double refraction. The O-ray is plane polarized with
vibrations perpendicular to the principal section and the E-ray is plane polarised with vibrations
parallel to the principal section. Travelling further, both the rays meet the Canada balsam layer along
AC.
The O-ray travels from denser (calcite) to rarer (Canada Balsam) medium. The dimensions of the
crystal are so chosen that angle of incidence of the O-ray at the calcite-balsam surface becomes
greater than the corresponding critical angle for the O-ray. Under these conditions, the O-ray is
totally reflected at the calcite-balsam surface into the calcite and is absorbed by the tube containing
the Nicol prism, as the inner surface of the tube is blackened.

The E-ray travels from rarer (Calcite) to denser medium (Canada balsam) and emerges from the Nicol
prism.

Thus, the light emerging from the Nicol prism is plane polarised with vibrations parallel to the
principal section.

Production of elliptically polarised light:

Let unpolarized light is incident on a Nicol prism. The emergent light is plane polarised. This plane
polarized light is incident on a quarter wave plate. The vibrations of this plane polarized light should
make an angle other than 0⁰,45⁰ & 90⁰ with the optic axis of the quarter wave plate.

NIST
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Solutions
The plane polarized light incident on the quarter wave plate splits into two polarized beams with
electric fields in mutually perpendicular directions. The amplitudes of these two beams are-

= , = in

=Amplitude of the electric field of the plane polarized light incident on the quarter wave plate.

=Angle of inclination of the vibrations of plane polarized light with the optic axis of the quarter
wave plate

Since, 0 , 0 , the two components have unequal amplitudes &, .

As the quarter wave plate introduces a phase differences of /2, the light coming out of the quarter
wave plate is elliptically polarized.
Production of circularly polarized light-

The Experimental arrangement used is the same as that used for production of elliptically
polarized light. The nicol prism is so rotated that the angle between the vibrations of plane polarized
light emerging out of the nicol prism and the optic axis of the quarter wave plate is 45⁰. So the
amplitudes of the two mutually perpendicular components of the electric field are equal.

Ex=E0 cosθ; Ey= E0 sinθ;

Sinθ=45⁰; Ex= Ey= E0/√2

As the quarter wave plate introduces a phase difference of /2,the light is coming out of the quarter
wave plate is circularly polarized.

5 (b) If the refractive indices for the ordinary ray in case of calcite and Canada balsam are 1.658
and 1.550respectively, calculate the maximum possible inclination with the Canada balsam
surface, so that the ordinary ray is still quenched. (3)

NIST
Soln:

If the said calcite acts as a Nicol prism then total internal reflection takes place at
canadabalsam surface.

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Solutions
For total internal reflection
2
sin ic  so maximum angle of incidence of ordinary light at Canada balsam surface is
1
   1.55 
ic  sin 1  2   sin 1    sin  0.93486   69.2
1 0

 1   1.658 
Hence maximum angle of inclination at Canada balsam surface is 900  690  210

Q 6.(a) Explain physical significance of grad ϕ (2)

Soln:

a) Physical significance of gradϕ :

 It is the directional derivative of the field represented by scalar function ϕ.
  When the value of ϕ is constant throughout a surface then grad ϕ directs along the
perpendicular drawn to the surface. This is why electric field intensity is normal to
equipotential surface.

Q 6.(b) Write Maxwell’s electromagnetic equations and discuss their physical significance. (6)

Soln:

Maxwell’s equations are

⃗ . ⃗ = ρ ………………………………(1)

⃗ . ⃗ = 0 ………………………………(2)

⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ + = 0 ………………… (3)

⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ -

NIST
= ……………………(4)

Physical significance:

 Maxwell’s equations incorporate all laws of electromagnetism.



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These equations are used to derive wave equation for electromagnetic waves.
 They are consistent with special theory of relativity.
 These equations are applicable for both classical and quantum e.m. field theory.
 These equations provide a unified description of both electric and magnetic
phenomena.

Solutions
Q 6.(c) State Ampere’s circuital law and write its differential form. (2)

Soln:

Statement: The line integral of magnetic field along a closed loop is equal to µ0
times the net electric current enclosed by the loop.
 c
B  dl  0 I net

Differential form: if j is the surface current density then I   j  d s

s

From Stoke’s theorem  c

 
B  dl    B  d s
s

putting these expressions in Ampere circuital law ,

   B   d s    j  d s
s
0
s


or    B  0 j  d s  0 
s

or   B  0 j
This is the differential form of Ampere circuital law.

Q 7.(a) What are the characteristics of electromagnetic waves ? (4)

Soln:

 It can travel in vacuum.

 It has two components. One is time varying electric field and other is time varying
magnetic field.
 Electric field and magnetic field are perpendicular to each other and also
perpendicular to the direction of propagation of the wave.
 It moves with the speed of light in free space.
 The oscillations of electric field & magnetic field in electromagnetic waves are in
phase with each other.

7 ( b) Derive electromagnetic wave equations for ⃗ and ⃗⃗ in a conducting medium. (6)

NIST
Soln:

In a charge free region volume charge density =0

In a conducting medium, electric current density = ⃗

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Where is the conductivity of the medium.

Maxwell’s electromagnetic equations are

⃗ . ⃗ = ρ => ε ⃗ . ⃗ = ρ ( as ⃗ = ε ⃗ ) ………………………………………………………………………..(1)

Solutions
⃗ . ⃗ = 0 ……………………………………………………………………………………………………………………..(2)

⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ + = 0 …………………………………………………………………………………………………..…….(3)

⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ - = => ⃗⃗⃗ ⃗ -με = μ ( as ⃗ = ε ⃗ & ⃗ = μ ⃗ ) ………………………(4)

If ϵ and μ are independent of position and time, then the Maxwell’s equation (1 ) to ( 4) in a charge
free conducting medium takes the form,

⃗ . ⃗ = 0 ………………………………………………………………………………………………………………………..(5)

⃗ . ⃗ = 0 ………………………………………………………………………………………………………………………..(6)

⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ + = 0 ……………………………………………………………………………………………………..…….(7)

⃗⃗⃗⃗⃗
⃗⃗⃗ ⃗ -με = μ ⃗ ……………………………………………………………………………………………..……(8)

Operating curl on both sides of Eq (7), we have

 ⃗⃗⃗⃗⃗
⃗⃗⃗ ( ⃗⃗⃗ ⃗ ) + ⃗⃗⃗ =0

 ⃗⃗⃗ (⃗ . ⃗ ) - ⃗ + ( ⃗⃗⃗ ⃗ )=0

⃗⃗⃗⃗⃗
 ⃗⃗⃗ (0) - ⃗ + [μ ⃗ +με ] = 0 [ by eq 5 & 8 ]
⃗⃗⃗⃗⃗ ⃗
 ⃗ -μ -με =0
⃗ ⃗⃗⃗⃗⃗
 ⃗ -με =μ …………………………………………… (9)

Similarly operating curl on both sides of Eq (8), we have

⃗⃗⃗⃗⃗
⃗⃗⃗ (⃗⃗⃗⃗⃗ ⃗ ) - ⃗⃗⃗ ( με ) = ⃗⃗⃗ ( μ ⃗ )

 ⃗⃗⃗ (⃗ . ⃗ ) - ⃗ -με ( ⃗⃗⃗ ⃗ )=μ ( ⃗⃗⃗ ⃗ )

⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
 ⃗⃗⃗ (0) - ⃗ με (- )=μ (- ) [ by eq 6 & 7 ]

NIST
⃗ ⃗⃗⃗⃗⃗
 ⃗ - με = μ ………………………………………….(10)

Equation (9) and (10) are the required wave equations for ⃗ and ⃗ in a conducting medium.

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8 (a) Show how uncertainty relation can be applied to Bohr’s orbit. (3)

Son:

Uncertainty principle is applied to Bohr’s orbit to find out the ground state energy of Hydrogen
atom.

Solutions
p2 e2
The energy of the Hydrogen atom is E  
2m 4 0 r
If p and r are the uncertainties in momentum and position respectively. Then by
Heisenberg’s uncertainty principle . ≈ or . ≈ which gives ≈ / .
2 2
e
Putting this in the expression for energy, E  
2mr 2
4 0 r
 dE   2 e2 
If the radius of ground state energy is r0,    0 or  3   0
 dr r0  mr 4 0 r 2  r  r
0

4 0 2 2
e2
Or r0  so energy at ground state is E0  
me2  4 0 2 
2
 4 0 2 
2m   4 0  2 
 me 
2
 me 
me4
Or E0  this matches with the ground state energy of Bohr’s orbit.
32 2 02 2

8 (b) Discuss physical significance of wave function Ψ (4)

Son:
  | Ψ |2 represents the probability density of finding a micro system in a
particular state. Hence Ψ is the probability amplitude of the system.
 The wave function Ψ satisfies Schrödinger’s both time dependent and
time independent wave equations.
 Ψ is single valued, analytic and continuous function obeying eigenvalue
equation. Hence Ψ is used to represent the system as well as its
different quantum states.
 Ψ is square integrable. Hence it can be normalised.

8 (c) Write Schrodinger time dependent wave equation and discuss its practical use. (3)

Son:

d 2
Schrödinger’s time dependent wave equation is i   2  V
dt 2m

NIST
It is used to describe various interactions of time varying fields like electric and magnetic fields in
electromagnetic waves. It is also used for investigating various quantum mechanical problems.

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Solutions