Engineering Physics Questions and Answers – Interference of Light

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Interference of Light”.

1. Which of the following does not show any interference pattern?
a) Soap bubble
b) Excessively thin film
c) A thick film
d) Wedge Shaped film
View Answer

Answer: b
Explanation: An excessively thin film shows no interference pattern because in that case, as the thickness of the film is negligible, the path difference, Δ, between the two reflected rays turns out to be λ/2 which is the condition of minima. Hence, the pattern is not observed.
The interference pattern is observed in the soap bubble, thick film, and wedge-shaped film.

2. The main principle used in Interference is _____________
a) Heisenberg’s Uncertainty Principle
b) Superposition Principle
c) Quantum Mechanics
d) Fermi Principle
View Answer

Answer: b
Explanation: Superposition principle is the basic principle used in the interference of light. When the incoming light waves superimpose constructively, the intensity increases while when they add destructively, it decreases.

3. When Two waves of same amplitude add constructively, the intensity becomes _____________
a) Double
b) Half
c) Four Times
d) One-Fourth
View Answer

Answer: c
Explanation: As we know, I ∝ A2. Thus, as the two waves add constructively, their amplitude becomes twice and hence the intensity becomes four times.

4. The shape of the fringes observed in interference is ____________
a) Straight
b) Circular
c) Hyperbolic
d) Elliptical
View Answer

Answer: c
Explanation: The fringes observed in an interference pattern are hyperbolic in shape. When the distance between the slits and the screen is large, they appear almost straight.

5. If instead of monochromatic light white light is used for interference of light, what would be the change in the observation?
a) The pattern will not be visible
b) The shape of the pattern will change from hyperbolic to circular
c) Colored fringes will be observed with a white bright fringe at the center
d) The bright and dark fringes will change position
View Answer

Answer: c
Explanation: When white light is used instead of monochromatic light, all the seven constituent colors produce their interference pattern. At the center of the screen, all the wavelengths meet in phase and, therefore, a white bright fringe is formed. Then the next fringe will be formed due to violet color as the wavelength is shortest for violet color. This will be followed by indigo, blue till red color.

6. Zero order fringe can be identified using ____________
a) White light
b) Yellow light
c) Achromatic light
d) Monochromatic light
View Answer

Answer: a
Explanation: When the white light is used, the central fringe is white in color while the rest are colored. Thus, the central fringe can be identified using white light.

7. Interference is observed only when the phase difference between the two waves is zero.
a) True
b) False
View Answer

Answer: b
Explanation: For interference pattern, the phase difference between the two rays must be constant. It is not necessary that the phase difference between the two rays has to be zero.

8. The shape of the pattern depends on the ____________
a) Distance between the slits
b) Distance between the slits and the screen
c) Wavelength of light
d) Shape of the slit
View Answer

Answer: d
Explanation: The shape of the interference pattern observed depends on the shape of the slits. If the shape of the slits changes, the shape of the fringes changes.

9. A thin sheet of refractive index 1.5 and thickness 1 cm is placed in the path of light. What is the path difference observed?
a) 0.003 m
b) 0.004 m
c) 0.005 m
d) 0.006 m
View Answer

Answer: c
Explanation: AS we know, the path difference introduced by the sheet = (μ – 1) t, where t is the thickness of the sheet.
Here, μ = 1.5 and t = 0.01 m
Therefore, Δx = 0.5 X 0.01 m
= 0.005 m.

10. According to stoke’s law, the expression for maxima is: 2μtcosr = ____________
a) nλ
b) 2nλ
c) (2n + 1) λ/2
d) (n + 1) λ/2
View Answer

Answer: c
Explanation: Stoke’s relation states that when a light ray is incident on the interface of the optically denser medium, then the reflected wave suffers a phase change of π. Thus, for maxima 2μtcosr = (2n + 1) λ/2 and for minima 2μtcosr = nλ.

11. The interference pattern of soap bubble changes continuously.
a) True
b) False
View Answer

Answer: a
Explanation: The top of the soap bubble is excessively thin, while the middle portion is thin and the bottom of the bubble is very thick. Thus, as the thickness varies, the interference pattern varies.

12. Which phenomenon is observed in the following figure?
Newton’s Rings phenomenon is observed in the following figure owing to the wedge
a) Wedge-Shaped film
b) Destructive Interference
c) Refraction
d) Newton’s Rings
View Answer

Answer: d
Explanation: In the figure, newton’s rings would be obtained. The circular fringes would be observed owing to the wedge shaped film on both sides of equal thickness.

13. When a thin plate of refractive index 1.5 is placed in the path of one of interfering beams of Michaelson Interferometer, a shift of 30 fringes is observed. If the thickness of plate is 0.018 mm, the wavelength of the used light is ___________
a) 4000 Å
b) 5000 Å
c) 6000 Å
d) 7000 Å
View Answer

Answer: c
Explanation: Here, μ = 1.5, n = 30, t = 1.8 X 10-5 m
Path difference due to the glass plate = 2(μ – 1) t
2(μ – 1) t = nλ
λ = 2(μ – 1) t/n
= 2 X 0.5 X 1.8 X 10-5/30
= 6000 Å.

14. A thin layer of colorless oil is spread over water in a container (μ = 1.4). If the light of wavelength 640 nm is absent in the reflected light, what is the minimum thickness of oil layer?
a) 179.6 nm
b) 198.3 nm
c) 207.6 nm
d) 214.3 nm
View Answer

Answer: d
Explanation: Let t be the thickness of the oil layer spread on water.
For minima, 2μtcosr = nλ
For minimum thickness, we take n = 1.
Therefore, tmin = λ/2μ
= 600 nm/2.8
= 214.3 nm.

15. In Newton’s ring experiment, the diameter of the 10th ring changes from 1.40 to 1.23 cm when a liquid is introduced between the lens and glass plate. What is the refractive index of the liquid?
a) 1.05
b) 1.15
c) 1.25
d) 1.35
View Answer

Answer: c
Explanation: We know, Dn2 = 4nλR/μ
Without liquid in air, μ = 1, D102 = 40λR
With liquid, D102’ = 40λR/μ
Dividing both we get, μ = D102 D102’
= {1.40/1.27}2
= 1.25.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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