B.

Sc DEGREE (CBCS) REGULAR EXAMINATIONS, APRIL 2024

Fourth Semester
Complementary Course - PH4CMT01 - PHYSICS-OPTICS & ELECTRICITY
(Common for B.Sc Mathematics Model I, B.Sc Statistics Model I)
2017 Admission Onwards (Time: 3 Hours Maximum Marks: 60)
Scheme of Evaluation for Question Paper Code : 2402104
Part A: Answer any ten questions.Each question carries 1 mark.
1. State the conditions on which two waves can produce interference pattern.
a) Monochromaticity
b) Coherence (For two conditions- 1 mark. One condition- 0.5 mark.)
2. Thick films illuminated by white light do not exhibit any color in white light. Explain why.
Interference in thin film will be observed if the thickness of the film is less than the coherence length of the
incident light waves. Normally, the coherence length of the light from ordinary sources is of the order of a
fraction of a millimeter. Therefore, interference is seen with the films of thickness of the order of a few
hundred microns only. It is because of this reason that thick films do not exhibit interference. (Any
explanation connecting thickness of the film and coherence length Or any explanation connecting
interference - 1 Mark ).
3. What is half period Element?
A Fresnel half period element or zone is a thin annular zone of the primary wavefront in which the
secondary waves from any two corresponding points of neighboring zones differ in path by λ/2. (Any
explanation on path difference of λ/2 -1 mark)
4. What is fraunhofer diffraction?
When the source of light and screen are at infinitely large distances from the obstacle, then the
diffraction pattern observed is called Fraunhofer diffraction. Can be easily observed in the laboratory by
placing the source and screen at the focal plane of two convex lenses. (Correct explanation- 1 mark)
5. In nicol prism though double refraction take place we could get a single beam of plane polarised
light. How?
Unpolarized light is made to fall on the crystal at an angle of about 15°. The ray after entering the
crystal suffers double refraction and splits up into o-ray and e-ray with differnt directions of vibrations. The
values of the refractive indices and the angles of incidence are such that the e-ray is transmitted while the
o-ray is internally reflected. The face where the o-ray is incident is blackened so that the o-ray is
completely absorbed. Then we get only the plane polarized e-ray coming out of the Nicol prism. (Out of
syllabus. Any reasonable attempt 1 Mark)
6. How can you show that both the ordinary and extraordinary rays are plane polarized with their
planes of vibration mutually perpendicular ?
An analyzer can be used to detect the plane polarized light. Plane polarized light after the double
refraction is passed through the analyser. Polarized lights with their mutually perpendicular planes of
vibration can be detected by the proper rotation of the analyser. (Any explanation analyser/its rotation).
7. What is optical activity?
Certain crystals and solutions possess a natural ability to rotate the plane of polarization about direction of
propagation. It is known as optical activity. (Correct defenition- 1 mark. )
8. What is population inversion?
Population inversion is the non-equilibrium condition of the material in which population of the upper
energy level 𝑁2 momentarily exceeds the population of the lower energy level 𝑁1. (Any explanation - 1
mark. )
9. What are the important characteristics of laser beam?
Monochromaticity, coherence, directionality and high output power (Any two points -1 mark )
10. Write down the relation between dielectric constant and temperature for ferroelectric crystals.
𝐶
Susceptibility χ = 𝑇−𝑇𝑐
C - Curie Constant, T - Temperature , 𝑇𝑐 - Curie Temperature

χ = ϵ𝑟 − 1 (Equation with terms explained -------------- 1 mark)

11. What is the energy supplied when a current of i units flows through an inductor of self inductance
L units?
When a current flows through an inductor with self-inductance (L), energy is stored in the magnetic field
within the coil. The energy supplied to the inductor during a change in current is given by
𝑡 𝐼
1 2
𝐸𝑛𝑒𝑟𝑔𝑦 = ∫ 𝑃𝑑𝑡 = ∫ 𝐿 𝑖 𝑑𝑖 = 2
𝐿𝐼 (Correct equation 1 mark)
𝑜 0

12. Show that L/R and RC both have the units of time.
𝑉 𝑉 𝑄 𝑉 𝑄
𝑅 = 𝐼
= (𝑄/𝑡)
𝑎𝑛𝑑 𝐶 = 𝑉
𝑅𝐶 = (𝑄/𝑡)
× 𝑉
= 𝑡, ( 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑡𝑖𝑚𝑒) (0.5
Marks)
𝑑𝐼
Voltage drop across inductor 𝑉 = 𝐿 𝑑𝑡
, comparing the units of quantities on both sides we get unit of L
𝑉. 𝑠 𝑉.𝑠 1
as 𝐴
. Unit of L/R is 𝐴
× (𝑉/𝐴)
= 𝑠 unit of time. (0.5 Marks)
(Any explanation about the time constants to be given 1 Mark)
Part B. Answer any six questions. Each question carries 5 marks.

13. A two slit young's experiment is done with monochromatic light of wavelength 600 nm.
Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a
transparent plate of thickness 0.5 mm is placed in front of one of the slit interference pattern shifts
by 5mm. Find the refractive index of the transparent plate.
Equation for lateral shift due to the the transparent sheet:
𝐷
∆𝑥 = (µ − 1 ) 𝑡 𝑑
where µ is the refractive index of the thin transparent plate and 𝑡 is the thickness of it.
Given λ = 600 𝑛𝑚, D = 10 cm, d = 5 mm, t = 0.5 mm
Substitution:
𝐷
∆𝑥 = (µ − 1 ) 𝑡 𝑑

1
 −2
−3 −3 10×10 30
5 × 10 = (µ − 1 ) × 0. 5 × 10 × −3 . Rearranging: µ = 25
= 1. 2
2×10
(Not part of the syllabus. Writing equation for Fringe width to be given 5 marks)

14. Two straight and narrow parallel slits one millimeter apart are illuminated by amonochromatic
light. Fringes formed on the screen held at a distance of 100 cm are 0.5 mm apart. What is the
wavelength of the light used?
λ𝐷 β𝑑
Fringe width β = 𝑑
→λ = 𝐷
(1 Mark)
−3 −3
Given β = 0. 5 𝑚𝑚 = 0. 5 × 10 𝑚, 𝑑 = 1 𝑚𝑚 = 1 × 10 𝑚 , 𝐷 = 100 𝑐𝑚 = 1 𝑚
(1 mark)
−3 −3
0.5 ×10 × 1 ×10 −6
λ= 1
= 0. 5 × 10 𝑚 = 500 𝑛𝑚 (Substitution and final answer with proper
units 3 marks.)
15. Newton’s rings are observed in reflected light of λ=5.9x10-5 cm. The diameter of 10th dark ring is
0.5 cm. Find the radius of curvature of the lens and thickness of the air film.
2
𝐷𝑛
Radius of curvature: 𝑅 = 4𝑛λ
where 𝐷𝑛 is the diameter of the nth ring. (1 mark)
−2 2 −4
(0.5×10 ) 0.25×10 2
Substituting 𝑅 = −7 = −6 = 0. 01059 × 10 𝑚 = 105. 9 𝑐𝑚 (2 marks)
4 × 10 ×5.9×10 23.6×10
2
Thickness of air film 𝑡 = 𝐷𝑛 / 8𝑅 (1 marks)
Substitution: 𝑡 = (0. 5 × 0. 5)/(8 × 105. 9) 𝑐𝑚 = 0. 000295 𝑐𝑚 = 2. 95 µ𝑚 (1 mark)
16. Obtain an expression for the dispersive power of a grating.
● The diffraction of the nth order principal maximum for a wavelength λ , is given by the equation:
(𝑎 + 𝑏) sin θ = 𝑛 λ………… with explanation of terms. (2 marks)
𝑑θ 𝑛 𝑛𝑁'
● Diff. the eqn.: (𝑎 + 𝑏) cos θ 𝑑θ = 𝑛𝑑 λ; 𝑑λ
= ( 𝑎+𝑏 ) cos θ
= cosθ
(3 marks)
17. Describe Polarization by scattering.
If a narrow beam of natural light incident on a transparent medium containing a suspension of
ultramicroscopic particles, the light scattered is partially polarized. The incident light causes electrons in
the scattering medium to vibrate. A vibrating electron emits most light in a direction perpendicular to its
vibration and none along the direction of its vibration. The electric field of the emitted radiation is parallel
to the direction of electron vibration. Hence light scattered through about 90° with respect to the incident
direction is strongly polarized. The direction of vibration of the E vector in the scattered light will be
perpendicular to the plane defined by the direction of propagation and the direction of observation, ie, the
plane of the paper, as shown in Figure.
------------------- Figure only (2 marks)
The light from a blue sky is quite strongly polarised,
particularly at 90°from the sun. It is not completely polarised
because a significant amount of sunlight has undergone
multiple-scattering, i.e. has been scattered more than once.

2
Light scattered twice through a total angle of 90°would be less polarised than light scattered once.
Explanation with figure (5 marks) - definition of polarisation only - 1 Mark

18. Write a note on (a) two different pump sources and (b) active medium in laser.
PUMPING:- For achieving and maintaining the condition of population inversion, we have to
continuously raise the atoms in the lower energy level to the upper energy level. It requires energy to be
supplied to the system. Pumping is the process of supplying energy to the laser medium with a view to
transfer it into the state of population inversion. Because N, is originally very much larger than N₂. a large
amount of input energy is required to momentarily increase N, to a value comparable to N₁ There are a
number of techniques for pumping a collection of atoms to an inverted state, Optical pumping, electrical
discharge and direct conversion are some of the methods of pumping. In optical pumping, a light source
such as a flash discharge tube is used to illuminate the active medium. This method is adopted in solid state
lasers. In the electrical discharge method, the electric field causes ionization of the medium and raises it to
the excited state. In semiconductor diode lasers, a direct conversion of electrical energy into light energy
takes place. --------------- 3 Marks (Pumping sources name only ------------1 mark)
ACTIVE MEDIUM:-Atoms are in general characterized by a large number of energy levels. However, all
types of atoms are not suitable for laser operation. Even in a medium consisting of different species of
atoms, only a small fraction of atoms of a particular type have an energy level system suitable for achieving
population inversion. Such atoms can produce more stimulated emission than spontaneous emission and
cause amplification of light. Those atoms, which cause laser action, are called active centers. The rest of
the medium acts as host and supports active centers. The medium hosting the active centers is called the
active medium. An active medium is a medium which when excited reaches the state of population
inversion and promotes stimulated emissions leading to light amplification.
-------------------------- 2 Marks
19. A dielectric material having dielectric constant 3 is placed in an electric field of intensity 105 v/m.
Find the polarization in the dielectric material.
Polarization 𝑃 = ϵ0χ𝐸 where χ is the susceptibility of the material and E is the applied electric field. (1
Mark)
Susceptibility χ = ϵ𝑟 − 1 where ϵ𝑟 is the dielectric constant. (1 Mark)
−12 2 −1 −2
ϵ0 = 8. 85 × 10 𝐶𝑁 𝑚 ; χ = ϵ𝑟 − 1 = 3 − 1 = 2 ( 1 Mark)
−12 −12 2
𝑃 = ϵ0χ𝐸 = 8. 85 × 10 × 2 × 105 = 1858. 5 × 10 𝐶/𝑚 (2 Marks) OR
5 −7 2
(Give marks if E is written as 10 𝑣/𝑚, and the value of P = 17.7x 10 𝐶/𝑚 ------- 2marks)
20. A 15 microFarad capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance
and the current(rms and peak) in the circuit. If the frequency is doubled, what happens to the
capacitive reactance and the current?
−6
Capacitive reactance, 𝑋𝑐= 1/2𝜋fC = 1/ (2x3.14x50x 15x10 ) = 212.3 Ω (1 mark)
R.M.S value of current 𝐼𝑟𝑚𝑠= V/𝑋𝑐 = 220/212.3=1.036 A ------------- (1 mark)

3
 Peak value of current, 𝐼0 = √2 . 𝐼𝑟𝑚𝑠 = √2 x 1.036 = 1.465 A ------------- 1 mark
When frequency doubled, ie f= 100 Hz,
−6
𝑋𝑐= 1/(2x3.14x100x 15x10 ) = 106.1Ω
𝐼𝑟𝑚𝑠= 220/106.1 =2.074 A
𝐼0 = √2 x 2.074 = 4.148 A ------------- 2 marks
or
When the frequency doubles, capacitive reactance is halved, results increased RMS and peak currents. (2
marks)
21. With necessary mathematical equations, Explain the dissipation of power when an ac is applied
to an LCR circuit.
Consider an ac circuit containing resistance, inductance and capacitance. E and I vary continuously with
time. Therefore power is calculated at any instant and then its mean is calculated over a complete cycle.
The instantaneous values of the voltage and current are given by
𝐸 = 𝐸0 sin ω𝑡 and 𝐼 = 𝐼0 sin(ω𝑡 − ϕ) 1 mark
𝜙 is the phase difference Hence power at any instant is between current and voltage. Power at any instant
is
1
𝑃 = 𝐸 × 𝐼 = 𝐸0𝐼0 sin ω𝑡 sin(ω𝑡 − ϕ) = 2
𝐸0𝐼0 [cos ϕ − cos(2 ω𝑡 − ϕ)] (1 Mark)
𝑇
∫𝐸 ×𝐼 𝑑𝑡
Average power consumed over one complete cycle is P = 0
𝑇 ----------------- 1 mark
∫𝑑𝑡
𝑜

Solving the above equation (steps) --------------------------- 1mark

1
𝑃 = 2
𝐸0 𝐼0 cos ϕ ie. 𝑃 = 𝐸𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 cos ϕ ( 1 mark )

Or True power = Apparent Power x power factor ------------------ 1 mark

Part C Answer any two questions. Each question carries 10 marks.

22. Derive an expression for conditions of brightness and darkness on a plane thin film.
(Reference: A Textbook of Optics - Dr. N. Subrahmanyam, Brijlal, Dr. N. Avadhanulu. )

Figure - 2 Marks
Geometrical Path difference:
𝐵𝐹 + 𝐹𝐷 − 𝐵𝐻 (1 Mark)
Optical Path Difference: Optical path difference
∆𝑎 = µ𝐿 = µ(𝐵𝐹 + 𝐹𝐷) − (𝐵𝐻)

Simplification using Snell's law : obtaining

optical path difference ∆𝑎 = 2 µ𝑡 cos 𝑟 (4
Marks)

4
Correction on account of Phase change at reflection (1 Mark)
Conditions for Maxima(Brightness) and Minima(Darkness) - (2 Marks)

23. Give the theory of a plane transmission grating and describe how it is used to determine the
wavelength of light using grating at normal incidence.

Considering the length of the answer: If theory of grating is clearly written with figure give 8 Marks
If experiment for measuring wavelength is completely written give 8 Marks
For Complete answer with theory and experiment - 10 Marks

24. With the help of geometry of optical fibre explain how light is propagated through and optical
fibre. Derive the equation of numerical aperture of an optical fibre.
Description of optical fibre. Explanation of Core, Cladding -
Explanation on total internal reflection and critical angle:
The refractive index of the core material 𝑛1must be slightly greater than that of the cladding 𝑛2.
At the core cladding interface the angle of incidence ϕ between the ray and the normal to the interface
𝑛2
must be greater than the critical angle ϕ𝑐 defined by : sin ϕ𝑐 = 𝑛1

The rays that are incident at the core clad interface at angles greater than the critical angle will propagate
through the fibre. (3 Marks )

Figure - 2 marks

2 2 −1 2 2
sin θ0 = 𝑛1 − 𝑛2 or θ0 = sin 𝑛1 − 𝑛2 .
2 2
Numerical Aperture is defined as the sine of the acceptance angle. 𝑁𝐴 = 𝑛1 − 𝑛2
(derivation 5 marks)

25. Discuss the current and voltage variations through a series LCR circuit when an AC is
applied to it. Discuss the conditions for resonance in that circuit. Also discuss the bandwidth and
sharpness of the circuit.

5
 Figure (1 Mak)
𝑑𝐼 𝑄
Total emf in the circuit is given by, 𝐿 𝑑𝑡
+ 𝐼𝑅 + 𝐶
= 𝐸0 sin ω𝑡 (1 mark)
2
𝑑𝐼 𝑑𝐼 1 𝑑𝑄
Differentiating with respect to t, 𝐿 2 + 𝑅 𝑑𝑡
+ 𝐶 𝑑𝑡
= 𝐸0 ω cos ω𝑡
𝑑𝑡
Trial solution of the form, 𝐼 = 𝐼0 sin ω𝑡 ---------------- 0.5 mark
Derivation of current in the circuit by substituting 𝐼 in above differential equation

----------------- 2.5 marks (equation only- 1 mark)

1 2
−1 𝜔𝐿− 𝜔𝐶 2
where ϕ = 𝑡𝑎𝑛 𝑅
and 𝑍 = 𝑅 + (𝑋𝑙 − 𝑋𝑐) is called impedance of the circuit.
1
At a particular frequency, 𝜔𝐿 = 𝜔𝐶
so that the impedance becomes minimum being given by
Z = R. This particular frequency 𝛎0 at which the impedance of the circuit becomes minimum and therefore
the current becomes maximum, is called the resonant frequency of the circuit. Such a circuit which admits
1
maximum current is called series resonant circuit. Thus at 𝛎0, we have 𝜔𝐿 = 𝜔𝐶
or
1 1
2𝜋 𝛎0𝐿 = 2𝜋 𝛎0𝐶
or 𝛎0 = ---------- 2 Marks
2𝜋 𝐿𝐶
𝐸0
The maximum current in the circuit= 𝐼0= 𝑅
. The variation of current with frequency of applied voltage is
shown in Figure.
The sharpness of resonance is defined as the rapidity with which the current falls from its resonant value (
𝐸0
𝑅
)with change in applied frequency.The sharpness of peak depends upon the resistance R of the circuit.

For low resistance, the peak is sharp.------------ 1 mark

------------------------- Figure (1 mark)
The band width is defined as range of frequencies specified
between two points on either side of resonant frequency
1
where current falls to times or power falls to half of its
2
value at resonance. Band width, BW= 𝜔2 − 𝜔1= R/L or
BW = 𝛎2 − 𝛎1= R/ 2𝜋𝐿 ----------- 1 mark
(Give Marks for Obtaining the answer using Phasor
diagrams)