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Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 1
SECTION- A
Short type question answer (50-75 words)
1. Attempt all parts. All parts carry equal
marks: (2x10=20)
A. Write the postulates of Special theory
of relativity.
Ans.
There are two postulates of special theory of
relativity.
(i)Laws of physics are same in all inertial
frame of references.
(ii) The speed of light in free space is same in
all inertial frame of reference.
B. What are massless particles?
Ans.
According to the theory of relativity, mass (m)
of any particle varies with velocity (v) of this
object by the following relation
2
2
1c
v
mm o
Where, mo is rest mass of the particle and c is
velocity of light.
2
2
1c
vmmo
If v=c, then
0
1112
2
o
o
m
mc
cmm
If any particle is moving with velocity of light,
then rest mass of the particle is zero. These
particles are called photons.
C. What is Displacement current?
Ans.
Current produced by changing electric field is
known as displacement current (Id). Idea of
displacement current is originated from the
parallel plate capacitor. Consider, a parallel
plate capacitor of plate area A
Electric field inside capacitor
DAQ
EDA
QD
A
QE
A
QE
o
o
o
Current due to this charge,
dsJdsdt
dDI
waygeneralIn
Adt
dD
dt
DAd
dt
dQI
dd
d
..
.)(
Jd is known as displacement current density.
D. Define Proper Length and Proper
Time.
Ans.
Proper Length- The length of an object in its
rest frame is called proper length and
represented by Lo. The length of any object in
motion with respect to an observed always
appears to the observer to be shorter than its
proper length.
Proper Time- The time which is determined
by events that occur at the same place in
observer‟s frame of reference is called proper
time.
E. In diffraction grating spectrum
,which orders of interference will not
be observed if opaque space is double
the slit widths.
Ans.-
Position of principal maxima in a grating
spectrum is given by
)1(sin nde
Where, n is the order of the maximum
Position of minima in the pattern is given
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 2
......3,2,1,
)2(sin
mwhere
me
If both conditions are simultaneously satisfied,
a particular order of maximum will be absent
in grating spectrum, these are known as absent
spectra.
Dividing equation (1) by (2)
)3(
sin
sin
m
n
e
de
m
n
e
de
As per question, opaque space is double of slit
widths i.e. d=2e
....9,6,3
3
3
2
mn
m
n
e
e
m
n
e
ee
Hence, 3rd
, 6th
and 9th
.. order spectra will be
absent.
F. What is specific rotation?
Ans.
Specific rotation for given wavelength of light
at a given temperature is defined as the rotation
produced by one decimeter of the solution
containing one gram of optically active
substances per cc solution.
If θ is the rotation produced by l decimeter
length of solution and C is concentration in
gram per cc, then specific rotation (S) is given
by
lCS
G. What is quarter wave plate?
Ans.
It is crystal plate of double referacting material
of a uniaxial crystal cut with its optical axis
parallel to the refracting faces. If thickness of
the plate is such that it may produce a path
difference of λ/4 between the O-ray and E-ray
when monochromatic plane polarized light of
wavelength λ is incident normal to the plate,
then plate is known as quarter wave plate.
If t is thickness of the plate and μE,μo are the
refractive index corresponding to E-ray and O-
ray, then for calcite crystal,
Eo
t
4
Similarly for quartz crystal
oE
t
4
H. Define population inversion.
Ans.
The term population inversion describes an
assembly of atoms in which the majority are in
energy levels above the ground state.
If N1 and N2 are the number of atoms in
ground state and higher state, then for
population inversion
N2>N1
This is a non equilibrium state and generally
forbidden in the systems as it is the tendency of
atoms to occupy the state of minimum energy
i.e. ground state.
I. How numerical aperture is related
with acceptance angle?
Ans.
Numerical aperture (NA) is a measure of the
amount of light that can be accepted by a fiber.
It is defined as the sine of acceptance angle.
This angle is measure of the light gathering
power of the fiber. If θm is the acceptance
angle, then 2
2
2
1sin nnNA m
Where, n1 and n2 are the refractive indices of
core and cladding
J. Give the main conditions of
interference.
Ans.
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 3
Main conditions of interference are as follows
(i)The interfering waves should be coherent i.e.
phase difference between them must remain
constant with time.
(ii)The two interfering sources should emit
light of same frequency
(iii)In case of polarized light, the two waves
should be in same state of polarization.
SECTION- B
2. Attempt any five parts. All parts carry
equal marks: (5x5=25)
(a)What must be the minimum number of
lines per cm in a half width grating to
resolve the wavelength 5890A0 and 5896 A
0.
Ans.
Given
A
Ad
ndN
cmperlinesofnumberN
spectrumofordern
nNd
gratingofpowersolving
AA
58932
58905896
2
658905896~
,1
,Re
5890,5890
21
21
21
So, number of lines per cm
9826
5893N
a) Light of wavelength 6000A0 falls
normally on a thin wedge shaped film of
refractive index 1.4 forming fringes that
are 2.0 mm apart. Find the angle of the
wedge
Ans.
For normal incidence, fringe-width is given
by
β = λ/(2μ. θ)
2
Given, μ=1.4,λ=6000Å=6000×10-10
m,
λ =6×10-7
m
β=2.0 mm=2×10-3
m
4
44
4
4
3
7
1063
1014.3
198
14.3
180101.1
101.1
6.5
106
1024.12
106
rad
b) Assuming that all the energy from a
1000 watt lamp is radiated
uniformly,calculate the average values of
the intensities of electric and magnetic
fields of radiation at a distance of 2m from
the lamp.
Ans.
P=1000 Watt=1000J/s
For uniform distribution of area, the surface
area will be given by
A= 4πr2=4π.2
2=16π m
2
Since, energy flow per unit area per second is
known pointing vector (S). So, magnitude of
this vector will be given by
16
1000
sec
S
Area
perEnergyS
According to the definition of poynting vector,
HES
In electromagnetic waves electric field vector
and magnetic field vector are perpendicular to
each other. So, magnitude of S will be given by
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 4
)1(16
1000
16
100090sin
EH
EHEHHES o
Intrinsic impedance of free space is given by
)2(6.376
6.376
EH
H
E
Putting this value in eq (1)
mvoltE
E
E
EE
/51.86
02.7496
24.50
0.376600
16
0.376600
6.37616
1000
6.37616
1000
16
1000
6.376
2
From eq. (2)
mAH
EH
/23.0
6.376
51.86
6.376
c) A circular lamina moves with its
plane parallel to the X-Y plane of a
reference frame S at rest. Assuming its
motion to be along the axis of X. Calculate
the velocity at which its surface area would
appear to be reduced to half to an observer
in frame S at rest.
Ans.
If r and D are the radius and diameter of
circular lamina, its surface area (S) is given by
42
22
2 DDrS
Since, lamina is moving in X-direction, so its
diameter in X-direction will be contracted by
2
22
2
2
14
2.
2.
.min.
,
1
c
vD
DyD
radiusmajorradiusorellipseofArea
DD
directiony
insamebewilldiameteritsHowever
c
vDD
x
y
x
According to problem
Area of ellipse=(1/2).area of circle
smv
cv
ccv
cv
c
v
c
v
c
v
D
c
vD
/106.2
866.0
2
732.1
2
3
4
3
4
3
4
11
4
11
2
11
4.
2
11
4
8
22
2
2
2
2
2
2
2
2
22
d) Determine the specific rotation of the
given sample of the sugar solution if the
plane of polarization is turned through
13.20.The length of the tube containing
10% sugar solution is 20cm.
Ans.
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 5
dcmcml
CGiven
lCS
220
1.0100
10%10,2.13,
11 )/()(662
132
1.02
2.13
ccgmdcmlc
S
e) A parallel beam of sodium light
(λ=5890A0)strikes a film of oil floating on
water .When viewed at an angle of 300 from
the normal 8th
dark band is seen
.Determine the thickness of the film
(refractive index of oil = 1.5)
Ans.
Condition for dark fringe of light reflected
from a film of thickness t is
2
35.1
1058905890,8,
)1(cos2
10
mAnGiven
nrt
From Snell‟s law
3
8.2cos
9
8
9
11
sin1cos
3
1
5.1
5.0sin
5.1
30sinsinsin
sin
sin
2
r
rr
r
n
ir
r
i
From eq (1)
8.2
1058908
10589083
8.2
2
32
10
10
t
t
mt
mmt
mt
6659.1
106659.11016659
1058908.2
610
10
f) The value of µe and µo for quartz is
1.5508 and 1.5418 respectively .Calculate
the phase retardation for λ = 5000A0 when
the plate thickness is 0.032 mm.
Ans.
For a quartz crystal, path difference between O
and E-ray will be given by
tOE .
Phase difference between these two rays or
phase retardation
tOE .2
.2
A
Ammt
Given OE
5000
10032.0032.0
,5418.1,5508.1,
7
61728.3nretardatio Phase
5
1080864.1
5
10032.0914.32
5000
10032.010914.32
5000
10032.0009.014.32
10032.0)5418.15508.1(5000
14.32
.2
nretardatio Phase
73
7
7
tOE
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 6
SECTION-C
Note: Attempt all questions. All questions
carry equal marks. (5x7=35)
1. Attempt any one part of the following:
a) Show that the apparent length of a
rigid body in the direction of its motion with
uniform velocity v is reduced by the factor
√1-v2/c2. Discuss the result.
Ans.
Consider, a rod is moving with velocity (v) in
x-direction. The length of this rod in the frame
of references (S), which is at rest is L. If x1
and x2 are position of ends of the rods in this
frame of reference is L, then
L = x2-x1
(1)
Consider, another frame of reference (S‟),
which is moving with the velocity of rod i.e. v.
Let the position of ends of this rod in this
frame be x’2 and x’1 . If L‟ is length of rod in
this frame of reference, then
L‟ = x’2-x’1
(2)
Since, relative velocity of the rod in S‟ frame is
zero, hence length of the rod in this frame of
reference will be proper length (Lo) of the rod.
i.e.
L‟ =Lo= x’2-x’1
(3)
Lorentz transformation is given by
22
1'
1
1 cv
vtxx
22
2'
2
1 cv
vtxx
From equation (3)
)4(1
1
)1(
1
11
22
22
22
12
22
1
22
2
cvLL
cv
LL
equationFrom
cv
xxL
cv
vtx
cv
vtxL
o
o
o
o
Since, v<c
So, L<Lo
Thus the length of the rod seems to be
contracted by a factor of 221 cv .
Discussion:
There may be three cases
(i) if v<<c
From equation (4)
o
o
LL
c
v
cvLL
1
12/122
If the velocity of rod is very much small
compared to velocity of light, then length of
rod will be same.
(ii) if v is comparable to c, then
L<Lo
(iii) if v=c
0
11
1
2/1
2/122
L
LL
ccLL
o
o
This is impossible. Hence, no object can never
attain the velocity of light.
b) Deduce the relativistic velocity
addition theorem .Show that it is consistent
with Einstein’s second postulate Ans.
Velocity addition theorm
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 7
Consider, a particle is moving with If any
particle in moving in X-direction with velocity
V, then Lorentz transformation for this particle
may be written as
22
2
22
'
1'
'
'
1
cv
c
vxt
t
zz
yy
cv
vtxx
(1)
(x, y, z, t) and (x‟, y‟, z‟, t‟) are the coordinates
of points in S and S‟ frame. v is the velocity of
S‟ frame in X-direction.
Components of velocity of particle in X, Y and
Z-direction Vx, Vy, and Vz in S-frame.
Corresponding components are Vx‟, Vy
‟ and
Vz‟ in S‟ frame.
By the definition of velocity,
'
''
'
''
'
'' .
,
dt
dzVand
dt
dyV
dt
dxV
dt
dzVand
dt
dyV
dt
dxV
xyx
zyx
(2)
Differentiating 1 with respect to t
22
2
'
22
'
1
'
'
1
cv
c
vxt
t
zz
yy
cv
vtxx
22
2
''
''
22
'
'
22
'
'
1
'
'
1
1
cv
c
vxt
t
dt
dz
dt
dz
dt
dy
dt
dy
cv
vV
dt
dx
cv
vdt
dx
dt
dx
x
2.Attempt any one part of the following:
a) Explain the concept of Maxwell’s
displacement current and show how it lead
to the modification of Ampere’s law.
Ans.
Displacement Current
According to Maxwell, a changing electric
field is equivalent to a current which flows as
long as electric field is changing. This
equivalent current produces the same magnetic
effect as an ordinary current in a conductor.
This equivalent current is known as
displacement current.
The idea of displacement current was
originated from the study of discharge of a
condenser. During the charging of a parallel
plate capacitor charge starts accumulating
gradually on the plate of capacitor and
accordingly current start decreasing in the
circuit. When the capacitor is fully charged,
current stops. The is no actual flow of charge
between the space of plates during charging. If
a compass is placed between the plates, the
needle deflects. This indicates that there is a
magnetic field between the plates though there
is no source of magnetic field in the gap. The
other source is nothing but the changing
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 8
electric field between the plates and is
equivalent to displacement current.
Consider, the circuit element shown in Figure
which consists of two wires connected to the
plates of a parallel plate capacitor. If a current
Ic flows in the wires, then the charge on the
plates of the capacitor
dt
dQIc
From ampere‟s circuital law
L
cILdB .
i.e. line integral of B around a closed path
equals to the conduction currents passes
through a surface S bounded by L.
However, in the above circuit, there are two
choices of surface S.
(i) S1 passing through one of the wires and
hence gives a contribution Ic to the circuital
law. i.e.
L
cILdB . for S1
(ii) S2 passes through the space between
the capacitor plates. It does not pass through
any conductor and hence does not give a
contribution to circuital law.
L
LdB 0. for S2
b) Show that the velocity of plane
electromagnetic waves in free space is given
by c = 1/√µ0 ε0
Ans.
Maxwell‟s equations are given by
t
EJB
t
BE
B
E
oo
o
0
For free space
000 J
)4(
)3(
)2(0
)1(0
t
EB
t
BE
B
E
oo
Taking curl of eq 3
)5(
,0
)4(&)1(
2
22
2
22
2
2
t
EE
t
EE
t
E
tE
eqeqFrom
t
BEE
t
BE
oo
oo
oo
Similarly by taking curl of eq 4
)6(2
22
t
BB oo
Equation (5) and (6) are the general wave
equation.
General wave equation is given by
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 9
)7(1
2
2
2
2
t
u
vu
Comparing eq(5), eq(6) and eq(7)
oo
oo
oo
v
v
v
1
1
1
2
2
lightofvelocitysmv
v
v
mFNA
oo
oo
/103
109
1
1
1094
1104
11
.1094
1,104
8
16
9
7
1
9
27
Hence, the velocity of electromagnetic waves
in free space is equal to the velocity of light.
3.Attempt any one part of the following:
a) . Discuss the phenomenon of
interference of light due to thin films and
find the condition of maxima and minima.
Show that the interference patterns of
reflected and transmitted monochromatic
source of light are complimentary
Ans.
Consider, a parallel plate transparent film of
thickness t and refractive index μ. A light
beam SA is entering into this film from air.
One part of this beam is reflected in direction
AR1 and other part is refracted in AB direction.
Beam AB is again reflected in BC direction
which is further refracted in direction CR2.
Two beams AR1 and CR2 reflected and
interfere.
Figure1: Interference in reflected light.
CD is perpendicular drawn from C to line
AR1. Now path difference (Δ) will be given by
Δ = Path ABC in film – Path AN in air
= μ (AB+BC)- AN (1)
From triangle ABM and BCM,
ir
rtAN
irtAN
iACAN
iAC
AN
sin.cos
sin.2
sin.tan2
sin
sin
In triangle ANC
rtCMAMAC
rBMCMAM
r
tBCAB
r
tBC
r
BMAB
rBC
BMr
AB
BM
tan2
tan
cos
2
coscos
cos,cos
From Snell‟s law
ri
r
i
sinsin
sin
sin
i.e.
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 10
r
rtAN
rr
rtAN
cos
sin2
sin.cos
sin.2
2
Putting values of AB, BC and AN in equation
1,
rt
rr
t
rr
t
r
rt
r
t
cos2
.cos.cos
2.
)sin1(cos
2.
cos
sin.2
cos
2.
2
2
2
Reflection at the surface of denser medium
gives additional path difference of λ/2 in ray
AR1, Now total path difference between two
rays
2cos2
rt
(i) Condition for maxima
For constructive interference,
Path difference =2 n.λ/2
i.e.
..2,1,0,2
).12(cos2
2cos2
2cos2
nnrt
nrt
nrt
This is condition for maxima.
When this condition satisfied, then the film
will appear in bright in the reflected light.
(ii) Condition for maxima
For Destructive interference,
Path difference should be odd multiple of λ/2
i.e.
....2,1,0,.cos2
2).112(cos2
22).12(cos2
2).12(
2cos2
nnrt
nrt
nrt
nrt
This is condition for minima.
When this condition is satisfied, then the film
will appear dark in the reflected light.
Interference in Transmitted light-
A light beam SA is entering into this film from
air. One part of this beam is refracted in AB
direction. Beam AB reflected in BC direction
which is refracted in direction CD. Refracted
part of AB and CD beam interfere and give
interefernce pattern
Similar to the phenomena of reflection, path
difference in transmitted may be calculated as
Δ = Path BCD in film – Path BL in air
=μ(BC+CD)-BL (1)
rt cos2
Condition of bright and dark fringes may be
given as
....2,1,0,).12(cos2 nnrt
These conditions are complementary to each
other.
..2,1,0,cos2 nnrt
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 11
b) Describe and explain the formation
of Newton’s rings in reflected
monochromatic light. Prove that the
diameter of bright rings are proportional to
the square roots of odd natural numbers.
A plano-convex lens of large radius of
curvature R is placed on a plane glass plate
with its curved surface downwards and is
illuminated from above with a parallel beam of
monochromatic light. Some of the light is
reflected from the upper surface of the glass
plate and some from the lower surface of the
lens; interference thus occurs by division of
amplitude, the fringes being localised in the air
gap between the lens and plate.
Condition for interference- Wedge shaped
film of air forms between the plano-convex
lens and glass plate. In this case, path
difference between two reflected rays will be
given by
2cos2
rt
For normal incidence, r=0
For air film, μ=1
Path difference will be given by the following
equation
22
t
Since, t is variable in this case, so there is need
to estimate this in terms of any fix value.
Since, radius of curvature R is fixed, so let us
calculate „t‟ in terms of „R‟.
...2
11
,,
)()(
,
,
2
22
22
R
rRR
rRRt
rQMRCQtMO
QMCQMO
CMQIn
RCO
MOCOt
2
22
1.2
2
1
.2
1
,2
11
2
2
2
2
2
2
R
r
R
rdifferencePath
R
rt
R
rRRRt
RrastermsorderhighergNegelectin
R
rRRt
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 12
(i) For bright fringes
Path difference should be an even multiple of
λ/2.
i. e.
2
)12(
2)12(
2)12(
22)2(
2)2(
2
2
2
2
Rnr
Rnr
Rnr
nR
r
nR
r
Diameter of bright fringes
.........3,2,1,)12(
tan2
2.)12(2).12(
2
)12(22
nnD
tconsR
RnRnD
RnrD
n
n
Thus the diameter of bright fringes is
proportional to the square-root of the odd
natural numbers.
Ratio of diameters corresponding to bright
fringes
646.2:236.2:732.1:1:::
7:5:3:1:::
4321
4321
DDDD
DDDD
Separation between successive rings =
0.732:0.504:0.410
Thus seperation between consecutive bright
rings decreases as the order increases.
(ii) For dark fringes
Path difference should be an odd multiple of
λ/2.
i. e.
....3,2,1,0,
.2
.2
.2
.22
)112(
22)12(
2)12(
2
2
2
2
nRnr
Rnr
RnRnr
nR
r
nR
r
Diameter of dark fringes
.........3,2,1,0,
tan4
2..4
22
nnD
tconsR
RnRnD
RnrD
n
n
Thus the diameter of bright fringes are
proportional to the square-root of natural
numbers.
Ratio of diameter of few dark fringes
2:732.1:414.1:1:0:::
4:3:2:1:0:::
4321
4321
DDDD
DDDD
Separation between successive rings =
1:0.414:0.318:0.268
Thus separation between consecutive dark
rings also decreases as the order increases.
Thus, central ringes are broader and peripheral
rings are closer.
3.Attempt any one part of the following:
a) Write short notes on
(i) Interference filters
Ans.
An intereference filter is a optical system that
transmits a very narrow range of wavelength
and provides a monochromatic beam of light.
They are composed of transparent glass or
quartz substrate on which multiple thin layers
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 13
of dielectric material, sometimes separated by
spacer layers
Permit great selectivity
Constructive and destructive interference
occurs between reflections from various
layers
Transmission determined by :
(i)thickness of the dielectric layers
(ii)number of these layers
(iii)angle of incident light on the filters
Advantage
(i)They can be used as reflectors in two and
three color analysis.
(ii)They usually do not themselves produce
fluorescence.
(iii)They are available in short pass versions.
(iv)They are excellent as primary barrier
filters.
Disadvantage
(i)Have lower blocking properties
(ii)Reduced passing properties
(iii)Their reflecting and passing properties are
not absolute, this should be considered while
dealing with multiple wavelengths
(ii) Antireflection coatings
An antireflective or antireflection (AR) coatin
g is a type of optical coating applied to the
surface of lenses and other optical devices to
reduce reflection.
This improves the efficiency of the system
since less light is lost. In complex systems
such as a telescope, the reduction in
reflections also improves the contrast of the
image by elimination of stray light.
Figure shows the phenomena of interference
that lead to antireflecting coating. In this the
refractive index of the coating material is
choosen that it has an intermediate value
between the glass and air.
For example that refractive index of MgF2 is
1.38.
The ray AB is incident on the system and at
point B it suffers reflection on the surface of
AR coating (denser medium) and proceeds
along BC. A part of it moves along BD in the
same medium and at D it suffers reflection
again at the surface of denser (glass) and
emerges out of along EF in air.
Hence, same phase change occurs in both the
rays at reflection.
Reflection at glass/AR coating and AR
coating/air bounaries will produce phase
change of π.
Path difference =2
.2
Total path difference between rays
rt
rt
rt
cos2
cos2
22cos2
Term „λ‟ may be omitted as addition and
subtraction of λ will not affect the path
difference.
For normal incidence,
Path difference = 2μ.t
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 14
Where, μ and t are refractive index and
thickness of AR coating material respectively.
For condition of destructive interference,
2μ.t= (2n+1).(λ/2),where, n=0,1,2..
For n=0
2μ.tmin = λ/2
tmin= λ/4 μ
Minimum thickness of the coating required
for no reflection at the centre of visible
spectrum (λ = 5.5×10-5
cm)
cmt
cmt
5
min
5
min
10966.0
38.14
105.5
4
b) Describe the construction and
principle of Nicol Prism. Explain its
working as a polarizer and analyser.
Ans.
The basic principle behind Nicol Prism is
based on its unique behaviour on the event of
incidence of light rays on its surface.
When an ordinary ray of light is passed
through a calcite crystal, it is broken up into
two rays:
• An „Ordinary ray‟ which is polarized and
has its vibrations perpendicular to the
principle section of the crystal and
• An extra-ordinary ray which is polarized
and whose vibration is parallel to the
principle section of the prism.
• If by some optical means, one of the two
rays eliminates, the ray emerging through
the crystal will be Plane polarized.
• In Nicol Prism, ordinary ray is eliminated
and Extra-ordinary ray, which is plane
polarized, is transmitted through the prism.
A calcite crystal‟s length is three times
its breadth. Let ADFGBC be such a
crystal having ABCD as a principle
section of the crystal with BAD = 700.
The end faces of the crystal are cut in
such a way that they make angles of
680 and 112
0 in the principle section
instead of 710 and 109
0. The crystal is
then cut into two pieces from one blunt
corner to the other along two pieces
from one blunt corner to the other along
a plane perpendicular to the extra
ordinary rays.
Refractive index of Calcite for O ray,
Refractive index of Canada balsam,
Refractive index Calcite of E ray,
Thus we see that the Canada Balsam is
optically denser than calcite for E ray
and rarer for O ray. Finally the crystal
is enclosed in a tube blackened inside.
Working as polariser
Working as an analyser
When principal section of both the Nicols are
parallel, then emitted E-ray from polarizer P
has vibrations parallel to principal section of
analyser A, so get freely transmitted through it.
In this setting of Nicols the intensity of emitted
light is maximum. This position and the
position when the angle between the principle
sections of two prisms is 180° is known as
“Parallel nicols
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 15
When Nicol A is rotated from its position,
intensity of light emitted from it decrease
and becomes zero when principal sections of
two planes at right angle to each other. In this
situation light emitted from polarising Nicol P
has vibration in a plane normal to principal
section of analysing Nicol A and is totally
internally reflected back from Canada Balsam
layer and no light is emitted. In this setting,
two Nicols are said to be “Crossed Nicols”.
5.Attempt any one part of the following:
a) Discuss the phenomenon of
Fraunhoffer diffraction at a single slit and
show that the relative intensities of the
successive maximum are nearly1: 4/9π2:4/25
π2:4/49 π2
Ans.
A slit is a rectangular aperture whose length is
large compared to its breadth. AB is a slit in
the figure shown below.
Let a parallel beam of monochromatic light of
wavelength λ be incident normally on upon a
narrow slit, AB=e. Let the diffracted light be
focused by a convex length L.
According to Huygen‟s theory, a plane wave
front is incident normally on the slit AB and
each point AB sends out secondary wavelets in
all directions. The rays preceding in the same
direction as the incident rays are focused O;
while those diffracted through an angle θ are
focused at point P.
Let the disturbance caused at P by the wavelet
from unit width of slit M be
tAy cos
Phase difference caused by the wavelet from
width dx at C when it reaches P,
sin.2
.2
x
encepathdiffer
Disturbance caused by this wavelet at P,
)sin.2
cos(.
)cos(.
xwtdxAdy
tdxAdy
For total disturbance
2
2
2
2
2
2
2
2
2
2
2
2
.sin.2
sin.sin
.sin.2
cos.cos
.
sin.2
sin.sin
sin.2
cos.cos
.)sin.2
cos(
)sin.2
cos(.
e
e
e
e
e
e
e
e
e
e
e
e
dxxwtA
dxxwtAy
dx
xwt
xwt
Ay
dxxwtAy
xwtdxAy
dyy
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 16
)()(0.sin.
.cossin2
.cos
sin2
..sin.sin
sin2
..cos.cos
sin2
..sin.sin
sin2
..cos.cos
sin2.,
sin2,
.sin.2
sin.sin
.sin.2
cos.cos
2sin2
2sin2
2sin2
2sin2
2sin2
2sin2
2sin2
2sin2
2sin2
2sin2
2
2
2
2
ufufwtA
duuwtAy
dutwtA
duuwtAy
dutwtA
duuwtAy
dudx
ux
let
dxxwtA
dxxwtAy
e
e
e
e
e
e
e
e
e
e
e
e
e
e
sinsin.
sin.cos
2
sin2sin2.
sin2.cos
2
sin2sin.2
sin2.cos
.cos2.cos
)()(
2sin2
0
2sin2
2sin2
2sin2
0
2sin2
2sin2
ewtAy
ewtAy
e
uduu
duuduu
ufuf
ee
e
ee
e
wtAwtAeyFor
wte
e
Aey
e
e
ewtAy
ewtAy
cos.cos.,0
cossin
sinsin
.
sin
sinsin
..cos
sinsin.
sin.cos
0
wtAy
e
cos.sin
.
sin
0
2
22
0
2
0
sinsin.
sin.
oIAI
AI
AA
For principle maxima
Intensity to be maximum,α should be zero.
ateInderthen
if
minsin
,0sin
,0
So, for estimation of maximum value of I,
2
2
max
sin
OLim
oII
For minima
sinα = 0 and
α= n, n = 1, 2, 3,…
ne
ne
e
sin
sin)/(
sin)/(
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 17
For secondary maxima
tansincos)(
0sin)(
0sincos
sin2
cos11
sinsin
2
sin
3
2
2
ii
or
i
I
Id
dI
II
O
O
O
The first condition (i) is just the condition for
minima that we have already seen (i.e. α = n,
n = 1, 2, 3, …)
The second condition (ii) results in a
transcendental equation whose graphical
solutions can be observed on the left. The
solutions can easily be found numerically: α
1.43, 2.46, 3.47…which approaches
(m+1/2) for large α.
c) Describe the method of production of
plane, circularly and elliptically polarized
light.
Production of plane polarized light-
When a beam of unpolarized light enters into a
nicol, it spilts up into ordinary ray and ordinary
ray
6.Attempt any one part of the
following:
a) Explain the action of He Ne Laser
.How is it superior to Ruby Laser.
A helium-neon laser, usually called a He-Ne
laser, is a type of small gas laser. HeNe lasers
have many industrial and scientific uses, and
are often used in laboratory demonstrations of
optics.
He-Ne laser is a four-level laser.
Its usual operation wavelength is 632.8 nm, in
the red portion of the visible spectrum.
It operates in Continuous Working (CW)
mode.
The setup consists of a discharge tube of length
80 cm and bore diameter of 1.5cm.
The gain medium of the laser, as suggested by
its name, is a mixture of helium and neon
gases, in a 5:1 to 20:1 ratio, contained at low
pressure (an average 50 Pa per cm of cavity
length ) in a glass envelope.
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 18
The energy or pump source of the laser is
provided by an electrical discharge of around
1000 volts through an anode and cathode at
each end of the glass tube. A current of 5 to
100 mA is typical for CW operation.
The optical cavity of the laser typically
consists of a plane, high-reflecting mirror at
one end of the laser tube, and a concave output
coupler mirror of approximately 1%
transmission at the other end.
HeNe lasers are normally small, with cavity
lengths of around 15 cm up to 0.5 m, and
optical output powers ranging from 1 mW to
100 mW.
The left side of the representation shows the
lower levels of the helium atoms.The energy
scale is interrupted and that there is a larger
difference in energy in the recombination
process than is evident in the diagram.
A characteristic of helium is that its first states
to be excited, 21S1 and 21S0 are metastable,
i.e. optical transitions to the ground state 11S0
are not allowed, because this would violate the
selection rules for optical transitions. As a
result of gas discharge, these states are
populated by electron collisions
A collision is called a collision of the second
type if one of the colliding bodies transfers
energy to the other so that a transition from the
previous energy state to the next higher or
lower takes place. Apart from the electron
collision of the second type there is also the
atomic collision of the second type. In the
latter, an excited helium atom reaches the
initial state because its energy has been used in
the excitation of a Ne atom. Both these
processes form the basis for the production of a
population inversion in the Ne system
Working
A description of the rather complex HeNe
excitation process can be given in terms of the
following four steps.
(a)When the power is switched on, An
energetic electron collisionally excites a He
atom to the state labeled 21So . A He atom in
this excited state is often written He*(21So),
where the asterisk means that the He atom is in
an excited state.
(b) The excited He*(21So) atom collides with
an unexcited Ne atom and the atoms exchange
internal energy, with an unexcited He atom and
excited Ne atom, written Ne*(3s2), resulting.
This energy exchange process occurs with high
probability only because of the accidental near
equality of the two excitation energies of the
two levels in these atoms. Thus, the purpose of
population inversion is fulfilled.
When the excited Ne atom passes from
metastable state(3s) to lower level(2p), it emits
photon of wavelength 632 nm.
This photon travels through the gas mixture
parallel to the axis of tube, it is reflected back
and forth by the mirror ends until it stimulates
an excited Ne atom and causes it to emit a
photon of 632nm with the stimulating photon.
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 19
The stimulated transition from (3s) level to
(2p) level is laser transition.
This process is continued and when a beam of
coherent radiation becomes sufficiently strong,
a portion of it escape through partially silvered
end.
The Ne atom passes to lower level 1s emitting
spontaneous emission. and finally the Ne atom
comes to ground state through collision with
tube wall and undergoes radiationless
transition.
b) Explain the various types of losses in
optical fiber.
Ans.
Losses in optical fiber-
7.Attempt any one part of the following:
a) Explain the phenomenon of double
refraction calcite crystal. Define ordinary
and extraordinary rays.
Ans. Double refraction calcite crystal-
Certain crystals like quartz, mica, calcite
have the property of producing two refracted
rays for
every ray that is incident on them.
The phenomenon of refraction where in two
refracted rays are produced for a given
incident ray
is called double refraction or birefringence.
The two refracted rays are plane polarized in
mutual
perpendicular planes. By eliminating one of
the them, plane polarized light can be
obtained. The
crystals which exhibit this property are called
doubly refracting crystals.
b) Describe a half shade polarimeter
and explain how it is used to measure the
strength of sugar solution
Polarimeter is an instrument used to measure the
specific rotation of an optically active solution.
Construction: It consists of two Nicol prisms N1
and N2 which acts as polarizer and analyzer
respectively. A glass tube T is placed between N1
and N2. A half shade device „H‟ is placed
between N1
and T. Half shade device is a circular plate, half
of which is quartz and the other half is glass. A
telescope is used to observe the light emerging
out of the analyzer. A circular main scale S and
two
verniers V1 and V2 are fixed to the analyzer.
Working:-
Light from a monochromatic source is made to
fall on polarizer N1 using a condensing lens. The
Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102
2012
Dr. Jitendra Pal Singh, KEC Ghaziabad Page 20
glass T is filled with distilled water. The light
emerging out of N1 is plane polarized. Half of
this beam
falls on quartz portion and the other half on the
glass portion of the half shade device. The plane
of
vibrations of light passing through glass portion
remains unaltered while that passing through
quartz
(Optically active) is rotated. Hence when viewed
through the telescope, we have two distinct
halves. By
rotating the analyzer both halves of field of view
are made equally bright. Reading R0 is noted.
Now,
the tube T is filled with optically active solution
of known concentration „c‟. When viewed
through the
telescope the brightness of the two halves are
different again. The analyzer is rotated to have
uniform
brightness again and the reading R is noted.
R~R0=θ gives the angle of rotation of the plane
of
polarization. The experiment is repeated for
different concentrations „c‟ and the results are
tabulated.