Diffraction of Light — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Diffraction of Light MCQs with step-by-step solutions (21 questions). Part of Wave Optics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Diffraction of Light · easy · theory
Diffraction of light is the phenomenon of:
A. Splitting of white light into colours
B. Restriction of vibrations to one plane
C. Reflection of light from a rough surface
D. Bending of light around the edges of an obstacle ✓ Correct
Solution: Light spreads into the geometrical shadow when it encounters an obstacle or aperture comparable in size to its wavelength.
Q2 — Diffraction of Light · medium · theory
In a single-slit diffraction pattern, the central maximum is:
A. Twice as wide as the secondary maxima ✓ Correct
B. Half as wide as the secondary maxima
C. Always dark
D. The same width as the secondary maxima
Solution: The central maximum spans from the first minimum on one side to the first on the other, covering two units of the spacing between successive minima.
Q3 — Diffraction of Light · medium · theory
The condition for minima in a single-slit Fraunhofer diffraction pattern of slit width $a$ is:
A. $a\sin\theta = n\lambda$ ✓ Correct
B. $a\cos\theta = n\lambda$
C. $a\sin\theta = \dfrac{n\lambda}{2}$
D. $a\sin\theta = (2n-1)\dfrac{\lambda}{2}$
Solution: The slit divides into pairs of strips whose contributions cancel exactly when the edge-to-edge path difference is a whole number of wavelengths.
Q4 — Diffraction of Light · easy · theory
Diffraction effects become most pronounced when the size of the obstacle or aperture is:
A. Comparable to the wavelength of light ✓ Correct
B. Much smaller than an atom
C. Independent of the wavelength
D. Much larger than the wavelength
Solution: This is why sound diffracts noticeably around a doorway while light, with its far shorter wavelength, casts sharp shadows.
Q5 — Diffraction of Light · medium · theory
Fraunhofer diffraction is observed when the source and the screen are:
A. Effectively at infinite distance from the slit ✓ Correct
B. In different media
C. Both very close to the slit
D. At unequal finite distances
Solution: Parallel incident and diffracted beams, usually arranged with lenses, define the Fraunhofer class; finite distances give Fresnel diffraction.
Q6 — Diffraction of Light · hard · theory
In a single-slit diffraction pattern, the intensity of the first secondary maximum compared with the central maximum is:
A. Half as large
B. Twice as large
C. Much smaller, about $4\%$ ✓ Correct
D. Equal
Solution: Most of the diffracted energy is concentrated in the central maximum, and the secondary maxima fade rapidly on either side.
Q7 — Diffraction of Light · easy · theory
A diffraction grating consists of:
A. A polished reflecting surface
B. Two slits separated by a small distance
C. A large number of equally spaced parallel slits ✓ Correct
D. A single very narrow slit
Solution: Thousands of slits per centimetre produce very sharp, widely separated maxima, which makes a grating an excellent spectroscopic tool.
Q8 — Diffraction of Light · easy · theory
The phenomenon of diffraction establishes that light:
A. Has a wave nature ✓ Correct
B. Travels in perfectly straight lines always
C. Consists only of particles
D. Carries no momentum
Solution: Bending into the geometrical shadow cannot be explained by rectilinear propagation of particles; it requires waves.
Q9 — Diffraction of Light · medium · numerical
In a single-slit diffraction pattern, the first minimum for light of wavelength $600\text{ nm}$ occurs at $30^\circ$. The slit width is:
A. $0.3\,\mu\text{m}$
B. $0.6\,\mu\text{m}$
C. $2.4\,\mu\text{m}$
D. $1.2\,\mu\text{m}$ ✓ Correct
Solution: $a\sin\theta = \lambda \Rightarrow a = \dfrac{600 \times 10^{-9}}{\sin 30^\circ} = \dfrac{6 \times 10^{-7}}{0.5} = 1.2\,\mu\text{m}$.
Q10 — Diffraction of Light · hard · numerical
A slit of width $0.1\text{ mm}$ is illuminated by light of wavelength $500\text{ nm}$ and the pattern is observed $1\text{ m}$ away. The width of the central maximum is:
A. $2.5\text{ mm}$
B. $10\text{ mm}$ ✓ Correct
C. $5\text{ mm}$
D. $20\text{ mm}$
Solution: Width $= \dfrac{2\lambda D}{a} = \dfrac{2 \times 5 \times 10^{-7} \times 1}{10^{-4}} = 10^{-2}\text{ m} = 10\text{ mm}$.
Q11 — Diffraction of Light · hard · numerical
Light of wavelength $500\text{ nm}$ falls on a slit of width $2\,\mu\text{m}$. The angle of the first diffraction minimum is approximately:
A. $7.2^\circ$
B. $14.5^\circ$ ✓ Correct
C. $30^\circ$
D. $45^\circ$
Solution: $\sin\theta = \dfrac{\lambda}{a} = \dfrac{5 \times 10^{-7}}{2 \times 10^{-6}} = 0.25$, so $\theta \approx 14.5^\circ$.
Q12 — Diffraction of Light · hard · numerical
The angular width of the central maximum in a single-slit pattern for $\lambda = 600\text{ nm}$ and $a = 0.6\text{ mm}$ is:
A. $2 \times 10^{-3}\text{ rad}$ ✓ Correct
B. $4 \times 10^{-3}\text{ rad}$
C. $10^{-3}\text{ rad}$
D. $10^{-4}\text{ rad}$
Solution: Angular width $= \dfrac{2\lambda}{a} = \dfrac{2 \times 6 \times 10^{-7}}{6 \times 10^{-4}} = 2 \times 10^{-3}\text{ rad}$.
Q13 — Diffraction of Light · hard · numerical
Light of wavelength $500\text{ nm}$ falls on a slit of width $1\,\mu\text{m}$. The second diffraction minimum occurs at an angle of:
A. $45^\circ$
B. $90^\circ$ ✓ Correct
C. $30^\circ$
D. $60^\circ$
Solution: $\sin\theta = \dfrac{2\lambda}{a} = \dfrac{2 \times 5 \times 10^{-7}}{10^{-6}} = 1$, so $\theta = 90^\circ$ — the minimum just grazes the slit plane.
Q14 — Diffraction of Light · medium · numerical
A grating of spacing $2\,\mu\text{m}$ is illuminated by light of $500\text{ nm}$. For the first order maximum, $\sin\theta$ equals:
A. $0.25$ ✓ Correct
B. $1.0$
C. $0.5$
D. $0.125$
Solution: $d\sin\theta = n\lambda \Rightarrow \sin\theta = \dfrac{5 \times 10^{-7}}{2 \times 10^{-6}} = 0.25$.
Q15 — Diffraction of Light · medium · numerical
The half angular width of the central maximum for $\lambda = 550\text{ nm}$ and slit width $1.1\text{ mm}$ is:
A. $5 \times 10^{-4}\text{ rad}$ ✓ Correct
B. $5 \times 10^{-3}\text{ rad}$
C. $10^{-3}\text{ rad}$
D. $2.5 \times 10^{-4}\text{ rad}$
Solution: Half angular width $= \dfrac{\lambda}{a} = \dfrac{5.5 \times 10^{-7}}{1.1 \times 10^{-3}} = 5 \times 10^{-4}\text{ rad}$.
Q16 — Diffraction of Light · hard · numerical
For $\lambda = 600\text{ nm}$, slit width $0.3\text{ mm}$ and screen distance $1.5\text{ m}$, the linear width of the central maximum is:
A. $6\text{ mm}$ ✓ Correct
B. $1.5\text{ mm}$
C. $3\text{ mm}$
D. $12\text{ mm}$
Solution: Width $= \dfrac{2\lambda D}{a} = \dfrac{2 \times 6 \times 10^{-7} \times 1.5}{3 \times 10^{-4}} = 6 \times 10^{-3}\text{ m}$.
Q17 — Diffraction of Light · hard · numerical
For a slit of width equal to three times the wavelength, the first diffraction minimum occurs at approximately:
A. $60^\circ$
B. $30^\circ$
C. $19.5^\circ$ ✓ Correct
D. $45^\circ$
Solution: $\sin\theta = \dfrac{\lambda}{a} = \dfrac{1}{3} \approx 0.333$, giving $\theta \approx 19.5^\circ$.
Q18 — Diffraction of Light · medium · numerical
A diffraction grating has $5000$ lines per centimetre. Its grating element is:
A. $2 \times 10^{-6}\text{ m}$ ✓ Correct
B. $5 \times 10^3\text{ m}$
C. $5 \times 10^{-6}\text{ m}$
D. $2 \times 10^{-4}\text{ m}$
Solution: $d = \dfrac{10^{-2}\text{ m}}{5000} = 2 \times 10^{-6}\text{ m}$.
Q19 — Diffraction of Light · medium · numerical
If the width of a single slit is doubled, the width of the central diffraction maximum:
A. Halves ✓ Correct
B. Doubles
C. Remains unchanged
D. Quadruples
Solution: Central width $= \dfrac{2\lambda D}{a} \propto \dfrac{1}{a}$, so a wider slit gives a narrower, brighter central band.
Q20 — Diffraction of Light · easy · numerical
If the wavelength of light is doubled, the width of the central diffraction maximum:
A. Remains unchanged
B. Halves
C. Doubles ✓ Correct
D. Quadruples
Solution: Central width $\propto \lambda$, so longer wavelengths diffract more strongly.
Q21 — Diffraction of Light · medium · numerical
Light of $400\text{ nm}$ falls on a slit of width $0.2\text{ mm}$. On a screen $1\text{ m}$ away, the first minimum lies at a distance of:
A. $0.5\text{ mm}$
B. $4\text{ mm}$
C. $2\text{ mm}$ ✓ Correct
D. $1\text{ mm}$
Solution: $y = \dfrac{\lambda D}{a} = \dfrac{4 \times 10^{-7} \times 1}{2 \times 10^{-4}} = 2 \times 10^{-3}\text{ m}$.