# Physics Questions and Answers – Wave Optics – Interference of Light Waves and Young’s Experiment

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Wave Optics – Interference of Light Waves and Young’s Experiment”.

1. In Young’s double-slit experiment with monochromatic light, how is fringe width affected, if the screen is moved closer to the slits?
a) Independent
b) Remains the same
c) Increases
d) Decreases
View Answer

Answer: c
Explanation: Fringe width is given as:
β = $$\frac {D\lambda }{d}$$
As the screen is moved closer to the two slits, the distance denoted by D decreases and so fringe width β increase.
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2. In Young’s double-slit experiment, lights of green, yellow, and orange colors are successively used. Write the fringe widths for the three colors in increasing order.
a) βG < βY < βO
b) βO < βY < βG
c) βO < βG < βY
d) βY < βG < βO
View Answer

Answer: b
Explanation: Fringe width is expressed as:
β = $$\frac {D\lambda }{d}$$.
Since the wavelength are related as shown ➔ λG < λY < λO, we can say that ➔ βG < βY < βO.

3. In Young’s double-slit experiment, the two parallel slits are made one millimeter apart and a screen is placed one meter away. What is the fringe separation when blue-green light of wavelength 500 nm is used?
a) 0.5 mm
b) 50 mm
c) 0.25 mm
d) 25 mm
View Answer

Answer: a
Explanation: Fringe width is given as:
β = $$\frac {D\lambda }{d}$$.
So β = $$\frac { (1 \times 500 \times 10^{-9} ) }{(10^{-3})}$$
β = 0.5 mm
Therefore, the fringe separation when blue-green light of wavelength 500 nm is used is 0.5 mm.
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4. What would be the resultant intensity at a point of destructive interference, if there are two identical coherent waves of intensity I0 producing an interference pattern?
a) 5 I0
b) 2 I0
c) I0
d) zero
View Answer

Answer: d
Explanation: Resultant intensity at the point of destructive interference will be as follows:
I = I0 + I0 + 2√I0 I0 cos 180o
I = 0
Therefore, the value of the resultant intensity at a point of destructive interference is zero.

5. What happens to the interference pattern if the phase difference between the two sources varies continuously?
a) Brightens
b) No change
c) Disappears
d) Monochromatic pattern
View Answer

Answer: c
Explanation: The positions of bright and dark fringes will change rapidly. Such rapid changes cannot be detected by our eyes. Uniform illumination is seen on the screen i.e., the interference pattern disappears.
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6. Two independent light sources act as coherent sources’.
a) True
b) False
View Answer

Answer: b
Explanation: Two independent sources of light cannot be coherent. This is because the light is emitted by individual atoms when they return to the ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

7. What will be the effect on the fringes formed in Young’s double-slit experiment if the apparatus is immersed in water?
a) Increases
b) Decreases
c) Remains the same
d) Independent
View Answer

Answer: b
Explanation: The wavelength of light in water (λ’ = $$\frac {\lambda }{\mu }$$) is less than that in air. When the apparatus is immersed in water, the fringe width decreases. This is the impact experienced by the apparatus when immersed in water.
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8. What would be the resultant intensity at a point of constructive interference, if there are two identical coherent waves of intensity I’ producing an interference pattern?
a) 4 I’
b) 0
c) I’
d) 2 I’
View Answer

Answer: a
Explanation: Resultant intensity at the point of constructive interference is given as follows:
I = I’ + I’ + 2√I’I’ cos 0o
I = 4 I
Therefore, the value of the resultant intensity at a point of constructive interference is 4 I’.

9. In Young’s double-slit experiment if the distance between two slits is halved and distance between the slits and the screen is doubled, then what will be the effect on fringe width?
a) Doubled
b) Decreases four times
c) Increases four times
d) Halved
View Answer

Answer: c
Explanation: Original fringe width is given as ➔ β = $$\frac {D\lambda }{d}$$.
So, the new fringe width is ➔ β’ = $$\frac {\lambda.2D}{ ( \frac {d}{2} ) }$$
β’ = 4β
Therefore, the new fringe width is 4 β.
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10. If the separation between the two slits is decreased in Young’s double-slit experiment keeping the screen position fixed, what will happen to the width of the fringe?
a) Decreases
b) Increases
c) Remains the same
d) Independent
View Answer

Answer: b
Explanation: Fringe width is expressed as:
β = $$\frac {D\lambda }{d}$$.
So, as the separation (d) between the two slits decreases, the fringe width increases. They are inversely proportional to each other.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter