UNIVERSITY OF PETROLEUM AND ENERGY STUDIES

Physics (SOE), Semester I

Worksheet –I (Unit 1)

Interference, Diffraction and Polarization

Key points and formulae

 Superposition of two coherent waves: resultant wave amplitude is given by:

o 𝐴𝑅 = [𝑎2 + 𝑎2 + 2𝑎1𝑎2cos(𝛿)]0.5
1 2
where, a1 and a2 are amplitudes, and δ is the phase difference.

 Resultant intensity: 𝐼𝑅 = 𝐼1 + 𝐼2 + 2√𝐼1𝐼2𝑐𝑜𝑠𝛿; where I1 and I2 are intensities of interfering

waves.
 Light of wavelength comparable to width of opening (λ ~ a) spreads out upon passing through
the opening. This phenomenon is called diffraction.
 If the source of light or the screen or both of them are at finite distances from the diffracting
aperture, the diffraction obtained is called Fresnel Diffraction. Fresnel diffraction is also called
near-field diffraction.
 If the source of light or the screen or both of them are effectively far enough from the aperture
so that they can be considered as at infinite distances from the aperture. This kind of diffraction
is called Fraunhofer Diffraction. This is also called far-field diffraction
 A single slit placed between a distant light source and a screen produces a diffraction pattern
which consists of
o A broad, intense central band, called the central maximum
o A series of less intense bands (as compared to central maxima) symmetrically placed
around the central maxima, called as secondary maxima.
o A series of dark bands symmetrically placed around the central maxima called minima.

The condition of minima for single slit is given by

𝑏𝑆𝑖𝑛𝜃 = 𝑛𝜆
b is the slit width and n is the order of the minima

Points of maximum intensity are given by

3 5 7
𝑏𝑆𝑖𝑛𝜃 ~ 𝜆 , 𝜆, 𝜆 … … … ..
2 2 2

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 For N-slit diffraction

Condition of Maxima is

(𝑎 + 𝑏)𝑆𝑖𝑛𝜃 = 𝑚𝜆 m = 0,1,2,3 ….
a+b is the grating element and m is the order of the maxima

Condition of minima is

𝑁(𝑎 + 𝑏)𝑆𝑖𝑛𝜃 = 𝑛𝜆

N is the number of slits and n is the order of minima and n = 1,2,3…….N-1, n # 0.N.2N,….

 A light wave is an electromagnetic wave in which electric & magnetic fields vibrate
perpendicular to the direction of wave propagation (Transverse wave).
 There are three types of polarized state of light: Plane polarized, circularly polarized and
elliptically polarized light.
 The light can be polarized by passing through the Polaroid filter (polarization by transmission).
In order to analyze this polarized light another polaroid filter is used, known as analyzer.
 The intensity of transmitted light depends on the angle θ between transmission axis and plane
of polarization of incident light:.

Expression for Malus Law

𝐼 = 𝐼0𝑐𝑜𝑠2𝜃

Expression for Brewester’s law

tan 𝑖𝑝 = 𝜇

𝜆
Thickness of Half wave plate is given by 𝑡= 2(𝜇𝑜 −𝜇𝑒 )

𝜆
Thickness of Quarter wav plate is given by 𝑡 = 4(𝜇𝑜 −𝜇𝑒 )

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 Multiple Choice Problems
Q1. Interference of Light shows
(a) Wave nature of light
(b) Particle nature of light
(c) Both (a) and (b)
(d) None of the above
Q2. Two Laser Beams of same wavelength and intensities 9I and I are superimposed. The
minimum and maximum intensities of the resultant beam are:
(a) 8I and 10I
(b) 0 and 10 I
(c) 4I and 16 I
(d) 4I and 10 I
Q3. The Phenomena of diffraction was discovered by
(a) Grimaldi
(b) Fresnel
(c) Fraunhoffer
(d) Rayleigh
Q4. The diffraction fringes obtained by a single slit are of
(a) Equal width and equal Intensities
(b) Unequal width and equal Intensities
(c) Equal width and unequal Intensities
(d) Unequal width and unequal Intensities
Q5. Polarization of light proves the
(a) Corpuscular nature of light
(b) Quantum nature of light
(c) Transverse nature of light
(d) Longitudinal nature of light

Q6. The polarizing directions of two polarizing sheets are parallel. Through what angle must
either sheet be turned so the intensity becomes one-half of initial value.
(a) 60° and 120°
(b) 45° and 135°
(c) 30° and 150°
(d) 0° and 180°

Q7. In Fraunhofer’s diffraction, incident light waves have type of wavefront.

(a) circular (b) spherical (c) cylindrical (d) plane.
Q8. In single-slit experiment, if the red colour is replaced by blue then .
(a) the diffraction pattern becomes narrower and crowded together
(b) the diffraction bands become wider
(c) the diffraction pattern does not change
(d) the diffraction pattern disappears.

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 Q9. In Fraunhofer’s diffraction the wavefront undergoing diffraction has to be
(a) cylindrical (b) spherical (c) elliptical (d) plane.
.
Q10. What are the characteristics of grating spectra?
(a) Spectral lines are almost straight and outer sharp.
(b) The spectral lines are more and more dispersed as order increases.
(c) It is situated symmetrically on both sides of zero order image.
(d) All the above.

Q11. Diffraction grating has a

(a) large number of equidistant slits
(b) only one slit
(c) large number of random distant slits
(d) circular slit.

Q12. In a diffraction grating, the condition for principal maxima is

(a) 𝑒 sin 𝜃 = 𝑛𝜆 (b) (𝑒 + 𝑑) sin 𝜃 = 𝑛𝜆
(c) 𝑑 sin 𝜃 = 𝑛𝜆 (d) sin 𝜃 = 𝑛𝜆.
Q13. Along the optic axis, O-ray and E-ray travel with
(a) same velocity (b) different velocities (c) both a & b (d) none.
Q14. Which of the following phenomenon demonstrates the transverse nature of light?
(a) Diffraction (b) polarisation (c) dispersion (d) interference.
Q15. Two plane polarised beams of equal amplitude having a phase difference of π/2, are
mutually perpendicular. The superposition of the two results in light.
(a) circularly polarized (b) plane polarized
(c) partially plane polarized (d) elliptically polarized.
Q16. The path difference introduced by half wave plate between ordinary and extraordinary
rays is
(a) λ/2 (b) λ/4 (c) λ/6 (d) λ/8.
Q17. The path difference introduced by a quarter wave plate between ordinary and
extraordinary rays is
(a) λ/2 (b) λ/4 (c) λ/6 (d) λ/8.

Tutorials

Q1. Two coherent beams produces interference pattern with resultant intensity of 17 units. If the
beams have individual intensities of 4 unit and 1 unit. Calculate the phase difference
between them. Ans. Cos-1(0.5)

Q2. Two coherent sources of intensity ratio α interfere. Prove that the visibility can be written as

2√𝛼
𝑽=
1+ 𝛼
Q3. Two coherent sources whose intensity ratio is 144:1 produce interference fringes. Deduce
the ratio of maximum to minimum intensity of the fringe system. [169:121]
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Q4. In a Young’s double-slit experiment, the slits are separated by 0.32mm and the screen is
placed 1.5m away. The distance between the central bright fringe an the fourth bright
fringe is measured to be 1.3 cm. Determine the wavelength of light.
Q5. Suppose Young’s experiment performed with blue-green light of wavelength 500 nm. The
slits are 1.2 mm apart and the viewing screen is 5.4 m from the slits. How far apart are the
bright fringes?
Q6. A diffraction grating has 5000 lines per cm and the total ruled width is 5 cm. Calculate for
a wavelength of 5000 Å in the second order (i) the resolving power and (ii) the smallest
difference in wavelength resolved. What happens if half the ruling width is covered?
[50000,0.1A]
Q7. A grating has 1000 lines ruled on it. In the region of wavelength = 6000 Å, find (i) the
separation between two wavelengths that can be just resolved in the first order spectrum
and (i) the resolving power in the second order. [6 A, 2000]
Q8. The distance between the first and fifth minima of a single slit diffraction pattern is .35mm
with the screen 40 cm away from the slit, using light of wavelength 550nm (a) find the slit
width (b) calculate the angle of the first diffraction minimum. [2.8 x 10-4 m, 1.96 x 10-3
radian]
Q9. Assume that the limits of the visible spectrum are arbitrarily chosen as 430nm and 680nm.
Calculate the number of rulings per millimeter of a grating that will spread the first order
spectrum through an angle of 20º.
Q10. A grating has 600 ruling/cm and is 6.0cm wide (a) What is the smallest wavelength interval
that can be resolved in third order at λ=5000Å (b) For this wavelength and this grating can
the resolution be improved. How?
Q11. Determine the minimum number of lines in a grating that are just able to resolve the
sodium lines of wavelengths 5890 and 5896 Angstroms in the first order spectrum.
Q.12. What angle is needed between the direction of polarized light and the axis of a polarizing
filter to reduce its intensity by 90.0%? Ans. 71.60
Q13. Quartz plate has μο = 1.54 and μE = 1.55 for sodium light. What minimum thickness of the
quartz plate between the crossed polarizer and analyzer will produce annulment of the light,
while the plate is cut parallel to optic axis.
Q14. A solution of camphor is alcohol in a tube of 200mm is found to give a rotation of 33ο of
the plane of vibration of the light passing through it .Calculate the concentration of the
solution. Specific rotation of camphor is = +54ο.
Q15. A slab of a particular material produces an angle of polarization of 55.50. A change in the
material leads to an increase of polarization by 20%. Calculate the refractive index of the
material.

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 LASERS

Key Points and formula

 LASER is an acronym for Light Amplification by Stimulated Emission of Radiation. LASER

is highly monochromatic, intense, collimated, directional, and very powerful source of light.

 The production of LASER light is based on the phenomenon of stimulated emission of

radiation.

 The number of atoms per unit volume (Population) in an energy level (E) at any temperature
(T) is given by Maxwell-Boltzmann distribution 𝑁 ∝ 𝑒−𝐸/𝐾𝐵𝑇, where KB is Boltzmann
constant.

 Population inversion is a necessary criterion for lasing action: if N2 and N1 are the number of
atoms in excited and ground states, respectively, then population inversion may be
𝑁
mathematically written as 2 > 1. If 𝐸2 and 𝐸1 are the energies of excited and ground states,
𝑁1
respectively, then 𝑁2 −(𝐸2−𝐸1)
= 𝑒𝑥𝑝 ( )
𝑁1 𝐾𝐵𝑇

 Relationship between Einstein’s A (spontaneous emission) and B (Stimulated Emission)

8𝜋ℎ𝜗3
coefficient is given by, 𝐴 = , where, h is Planck’s constant, 𝜗is frequency and c is
𝐵 𝑐3
velocity of light.

 Population inversion (and hence lasing) cannot be achieved in a two level system. Thus, a three level
and four level lasing schemes were employed.

 The pumping mechanism namely Optical pumping and Electrical pumping are used in LASER’s to
achieve population inversion.

Multiple choice Questions

Q1. The process in which photon emission occurs without any interaction with external radiation is
(a) Spontaneous emission

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 (b) Stimulates Emission
(c) Spontaneous absorption
(d) Stimulated absorption
Q2. The relation between coherence time Tc and coherence length Lc is:
(a) 𝐿𝑐 = 𝑐𝑇𝑐
(b) 𝐿𝑐 = 𝑐/𝑇𝑐
(c) 𝐿𝑐 = 𝑇𝑐/𝑐
(d) None of these
Q3. The line width of LASER beam of wavelength λ for a spread Δλ in wavelength will be:
(a) 𝜆 𝛥𝜆
(b) 𝛥𝜆/𝜆
(c) 𝜆/𝛥𝜆
(d) None of the above
Q4. The type of pumping used in Ruby laser is
(a) Chemical Pumping
(b) Electric discharge
(c) Optical pumping
(d) In-elastic atom-atom collisions
Q5. Population inversion is not possible in
(a) a 2-level pumping scheme
(b) a 3-level pumping scheme
(c) a 4-level pumping scheme
(d) none of the above
Q6. The He-Ne Laser is a
(a) 3-level Laser (b) 4-level Laser
(c) 2-level Laser (d) None of these

Q7. What color of light does the Ruby LASER emit?

(a) Blue (b) Yellow
(c) White (d) Red

Tutorial
Q1. Determine the MKS units of 𝑢(𝜈), 𝐴 𝑎𝑛𝑑 𝐵.
Ans: 𝑱𝒔𝒎−𝟑, 𝒔−𝟏, 𝒎𝟑𝑱−𝟏𝒔−𝟐

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Q2. Calculate the power per unit area delivered by a LASER pulse of energy 4×10-3 Joule and
pulse length as 10−9 sec when the pulse is focused on target to a very small spot of radius 0.15
µm. Ans: 5.6× 1019 W/m2

Q3. A LASER beam has a wavelength of 7200 Å with an aperture of 5.0 mm. The LASER beam
is sent on moon at a distance 4×108 m from earth. Determine its angular spread and areal
spread when it reaches moon. Ans: 1.75×10-4 radians; 4.9×108 m2

Q4. Find the relative population of the two states in a ruby laser that produces a light beam of
wavelength 6943 Å at 300 K and 500 K. Ans:𝟏. 𝟐𝟏 × 𝟏𝟎−𝟑𝟎 ; 𝟏. 𝟏𝟐 × 𝟏𝟎−𝟏𝟖

Q5. A pulsed laser is constructed with a ruby crystal as the active element. The ruby rod contains
typically a total of 3×1019of Cr+3 ions. If the Laser emits radiation of wavelength 6943 Å,
calculate the total energy available per Laser pulse (assuming total population inversion).
Ans: 8.6 J

Q6. For a light of frequency 1.5×1015 Hz used as an excitation source at a temperature of 5000 K,
find out the ratio of spontaneous and stimulated emission. Ans: 2.07×10-12 Jsm-3

Q7. Consider the yellow light (𝜈 = 5 × 1014𝐻𝑧) and T=2000 K, show that for optical region the
spontaneous emission dominates over the stimulated emission.

Q8. The wavelength of emission is 6000 Å and the lifetime τsp is 10-6 s. Determine the coefficient
for the stimulated emission. Ans:𝟏. 𝟑 × 𝟏𝟎𝟏𝟗𝒎/𝒌𝒈

Q9. Calculate the population ratio of two states in He-Ne Laser that produces light of wavelength
6000 Å at 300 K? Ans: ≈ 𝒆−𝟖𝟎

Q10. Calculate the ratio of stimulated to spontaneous emission at a temperature of 300 °C for
Sodium D line. Ans: 8.12×10-14

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Fiber Optics
 If 𝑛1 be the refractive index of the core and 𝑛2be the refractive index of the cladding, then
for total internal reflection

 𝑛1>𝑛2, i.e., the refractive index of core should be greater than cladding
 The angle of incidence on core-cladding interface (w.r.t. vertical) should be greater
than the critical angle.

 When 𝜃 > 𝜃𝐶 , where 𝜃 is the angle of incidence and 𝜃𝐶 is the critical angle, the ray
traverses along the interface (i.e. the angle of refraction is 90°) via total internal reflection.
𝑛 21−𝑛21
 Relative refractive index, ∆≡
2𝑛21

 Acceptance angle: It is the maximum angle w.r.t the axis of the optical fiber at which if
the light is sent to the optical fiber it will be guided through via total internal reflections:
 𝑖𝑚 = sin−1 (√𝑛2 − 𝑛2⁄𝑛0)
1 1

 Where, 𝑛0 is the refractive of the medium in which the optical fiber is kept. Acceptance
angle measures the light/signal−1gathering power of an optical fiber. Assuming the outside
medium is air 𝑛0 = 1, 𝑖𝑚 = sin √𝑛2 − 𝑛2 = 𝑛1√2∆
1 1

 Numerical aperture (NA): This is closely related to the acceptance angle of an optical
fiber. The more the acceptance angle is, more will be the numerical aperture:
 𝑁𝐴 = sin 𝑖𝑚 = √𝑛2 − 𝑛2, Numerical aperture determines the efficiency of coupling
1 1
from a source (generating the signals to be transmitted) to the fiber.

 Based on transmission properties and the structure, we can categorize optical fibers as a
single mode fiber or multimode fiber.

 Optical fibers may further be classified based on whether the refractive index of the core
is uniform or not. In step-index fibers, the refractive index of the core is uniform throughout
its diameter. In graded-index fibers, the refractive index varies across the diameter. Graded
index fibers are employed to minimize losses due to signal dispersion.

 The normalized waveguide parameter is known as V-number, given by:

2𝜋
𝑉 = 𝑎√𝑛2 − 𝑛2, where ‘a’ if radius of core and 𝜆 is wavelength.
𝜆0 1 1 0

 The step indexed optical fiber has only one mode if 𝑉 < 2.4048
𝑉2 𝑉2
 Number of modes in optical fiber is given by : 𝑁 = ⁄2 (SIF), 𝑁 = ⁄4 (PIF)

Multiple Choice Questions

Q1. The most commonly used material for glass fiber is
(a) B2O3 (b) P2O5 (c) GeO2 (d) SiO2

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Q2. The propagation (without attenuation) of light in optical fiber is based on the phenomenon of
(a) Reflection (b) Diffraction
(c) Refraction (d) Total internal reflection

Q3. The optical fiber whose core has a refractive index that decreases with increasing radial distance
from fiber axis is known as:
(a) Graded index fiber (b) Step index fiber
(b) Both (a) and (b) (d) none of these

Q4.The attenuation in optical fiber is caused primarily by:

(a) Absorption (b) scattering loss
(c) Bending loss (d) both (a) and (b)

Q5. The relation between numerical aperture NA and fractional refractive index change Δ is:
(a) 𝑁𝐴 = 𝜇1√2∆ (b) 𝑁𝐴 = 𝜇2√2∆
(c) 𝑁𝐴 = 2𝜇1∆ (d) 𝑁𝐴 = 2𝜇2∆

Q6. The optical power of 0.5 mW is initially launched into an optical fiber. The power level is found
to be 19.9 µW after 4 km. The attenuation in this fiber will be:
(a) 1.5 dB/km (b) 2.5 db/km
(c) 3.5 db/km (d) 345 db/km

Tutorial
Q1. A step index fiber has a numerical aperture of 0.21, core refractive index of 1.565 and a core
diameter of 100 micrometer. Calculate the refractive index of cladding. Ans: 1.5508

Q2. A light ray enters from air to step indexed multimode fiber, the refractive index of core is 𝑛1 = 1.5
and ∆= 0.015. Determine the refractive index of cladding, critical angle for propagation, NA and
maximum acceptance angle. Ans: 1.477, 79.95°, 0.26 and 15°

Q3. Repeat the above problem if light enters the optical fiber from water instead of air (µ=1.33).
Ans: 1.477, 79.95°, 0.26, and 11.27°
Q4.The distance between two successive reflections is known as skip distance (Ls) and number of
reflection in a given length (in meters) of fiber is Nr. Determine the skip distance and number of
reflections per meter for a fiber in which light is entering from water (µ=1.33) at an angle of 30°
with 𝑛1 = 1.6, and fiber diameter 50µm. Ans: 152 µm, 6580 reflections

Q5. Assume a step indexed fiber with diameter of 0.0025 inch. Has a core refractive index of 1.53
and cladding of index 1.39 (although the RI difference in very less in practice). Determine (a) the
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 numerical aperture, (b) the critical angle, (c) the number of reflections in a 3 ft fiber for a ray at
the maximum entrance angle, and (d) at one half of this angle.
Ans: (a) 0.64, (b) 79.5° and (c) 6624 and (d) 3281
Q6.An optical fiber cable 3 km long is made up of three 1km length fibers spliced together. The
losses due to each length and splice are 5dB and 1 dB respectively. What would be the output
power if the input power is 5 mW? Ans: 0.080 mW

Q7.A signal of 5 mW is send through the 100 m long fiber and the output power is 1 µW. What is
the attenuation coefficient (loss) of the fiber in dB/km? Ans: ~ 369dB/km

Q8.How many modes can propagate in the step indexed fiber with 𝑛1 = 1.461 𝑎𝑛𝑑 𝑛2 = 1.456 at
850 nm? The core radius is 20 µm. Ans: ~ 159

Q9. Consider a SIF with 𝑛1 = 1.5, 𝑎 = 40 µ𝑚 𝑎𝑛𝑑 ∆ = 0.015 operating at 850 nm with the spectral
width radius of 50 nm. Is this a single mode or multimode fiber?

Q10. Consider a AlGaAs optical fiber with 𝑛1 = 3.60 𝑎𝑛𝑑 𝑛2 = 3.55. How many modes can
propagate in this wave guide if diameter (d) = 0.5𝜆 and 5 𝜆. Ans: 1 and 44

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