# Physics Questions and Answers – Wave Optics – Diffraction

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Wave Optics – Diffraction”.

1. State the essential condition for diffraction of light to occur.
a) The size of the aperture must be less when compared to the wavelength of light
b) The size of the aperture must be more when compared to the wavelength of light
c) The size of the aperture must be comparable to the wavelength of light
d) The size of the aperture should not be compared to the wavelength of light

Explanation: Diffraction is the spreading out of waves as they pass through an aperture or around objects. The diffraction of light occurs when the size of the obstacle or the aperture is comparable to the wavelength of light.

2. What is the cause of diffraction?
a) Interference of primary wavelets
b) Interference of secondary wavelets
c) Reflection of primary wavelets
d) Reflection of secondary wavelets

Explanation: Diffraction occurs due to interference of secondary wavelets between different portions of a wavefront allowed to pass across a small aperture or obstacle. Interference can be either constructive or destructive. When interference is constructive, the intensity of the wave will increase.

3. What should be the order of the size of an obstacle or aperture for diffraction light?
a) Order of wavelength of light
b) Order of wavelength of obstacle
c) Order in ranges of micrometer
d) Order in ranges of nanometer

Explanation: The size of the obstacle or aperture should be of the order of the wavelength of light used. Therefore, the size of an obstacle is not of the order of obstacle, or micrometer or nanometer.
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4. A small circular disc is placed in the path of light from a distant source. Identify the nature of the fringe produced.
a) Dual
b) Narrow
c) Dark
d) Bright

Explanation: Waves from the distant source are diffracted by the edge of the disc. These diffracted waves interfere constructively at the center of the shadow and produce a bright fringe. Therefore, the nature of the fringe produced is bright.

5. Single slit diffraction is completely immersed in water without changing any other parameter. How is the width of the central maximum affected?
a) Insignificant
b) Increases
c) Decreases
d) Becomes zero

Explanation: Wavelength of light in water decreases, so the width of the central maximum also decreases. This is the impact on the width of the central maximum when a single slit is completely immersed in water.

6. Diffraction is common in light waves.
a) True
b) False

Explanation: As the wavelength of light is much smaller than the size of the objects around us, so diffraction of light is not easily seen. Therefore, for diffraction of a wave, an obstacle or aperture of the size of the wavelength of light of the wave is needed. As the wavelength of light is of the order of 10-6m and $$\frac {obstacle}{aperture}$$ of this size are rare, diffraction is not common in light waves.

7. Determine the half angular width of the central maximum, if a wavelength of 1000 nm is observed when diffraction occurs from a single slit of 2 μm width.
a) 100o
b) 30o
c) 90o
d) 150o

Explanation: Half angular width of the central maximum is given by:
Sin θ = $$\frac {\lambda }{d}$$
Sin θ = $$\frac {1000 \times 10^{-9}}{2 \times 10^{-6}}$$
Sin θ = 0.5.
Θ = 30o

8. What will be the linear width of the central maximum on a screen that is kept 5 m away from the slit, if a light of wavelength 800 nm strikes a slit of 5 mm width.
a) 1.2 mm
b) 5.6 mm
c) 6.5 mm
d) 9.7 mm

Explanation: Linear width is given as:
β0 = $$\frac {2D\lambda }{d}$$
β0 = $$\frac { ( 2 \times 5 \times 800 \times 10^{-9} ) }{ (5 \times 10^{-3} ) }$$
β0 = 1.6 × 10-3 m
β0 = 1.6 mm
Therefore, the linear width of central maximum on a screen kept 5 m away from the slit is 1.6 mm.

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