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Interference and Diffraction of Light — MH-CET Physics MCQs with Solutions
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Sample questions with solutions
Q1 — Interference · easy · theory
The phenomenon of interference of light is based on the principle of:
A. Quantisation of energy
B. Total internal reflection
C. Superposition of waves ✓ Correct
D. Rectilinear propagation
Solution: Interference results from the superposition of two (or more) coherent light waves.
Q2 — Interference · easy · theory
Two sources of light are said to be coherent if they have:
A. Different frequencies
B. The same amplitude only
C. A randomly varying phase difference
D. The same frequency and a constant phase difference ✓ Correct
Solution: Coherent sources emit waves of the same frequency with a constant (time-independent) phase difference.
Q3 — Interference · easy · theory
For constructive interference at a point, the path difference between the two waves must be:
A. $(2n+1)\lambda/4$
B. $n\lambda$ (n = 0, 1, 2, …) ✓ Correct
C. $(2n-1)\lambda/2$
D. $n\lambda/2$
Solution: Constructive interference (bright fringe) occurs when the path difference is an integral multiple of the wavelength, nλ.
Q4 — Interference · easy · theory
For destructive interference, the path difference between the two waves must be:
A. $2n\lambda$
B. $n\lambda$
C. $n\lambda/4$
D. $(2n-1)\dfrac{\lambda}{2}$ ✓ Correct
Solution: Destructive interference (dark fringe) occurs when the path difference is an odd multiple of λ/2.
Q5 — Interference · easy · theory
In Young's double-slit experiment, the fringe width β is given by:
A. $\beta = \dfrac{\lambda D}{d}$ ✓ Correct
B. $\beta = \dfrac{D d}{\lambda}$
C. $\beta = \dfrac{\lambda d}{D}$
D. $\beta = \dfrac{\lambda}{D d}$
Solution: Fringe width β = λD/d, where D is the slit-to-screen distance and d the slit separation.
Q6 — Interference · easy · theory
In Young's double-slit experiment, the central point of the screen (equidistant from both slits) is:
A. Alternately bright and dark
B. Uniformly grey
C. A bright fringe (zero path difference) ✓ Correct
D. A dark fringe
Solution: At the centre the path difference is zero, giving constructive interference — the central bright fringe.
Q7 — Interference · easy · theory
A path difference of one full wavelength (λ) between two interfering waves corresponds to a phase difference of:
A. 4π
B. π/2
C. π
D. 2π ✓ Correct
Solution: Phase difference = (2π/λ) × path difference = (2π/λ) × λ = 2π.
Q8 — Diffraction · easy · theory
Diffraction of light is the phenomenon of:
A. Bending of light around the edges of an obstacle or aperture ✓ Correct
B. Splitting of white light into colours
C. Rotation of the plane of vibration
D. Reflection at a polished surface
Solution: Diffraction is the bending/spreading of light as it passes the edges of an obstacle or a narrow aperture.
Q9 — Diffraction · easy · theory
Diffraction effects become significant when the size of the aperture or obstacle is:
A. Very much larger than the wavelength
B. Exactly one metre
C. Independent of the wavelength
D. Comparable to the wavelength of light ✓ Correct
Solution: Appreciable diffraction occurs only when the obstacle/aperture is of the order of the wavelength of light.
Q10 — Diffraction · easy · theory
For a single slit of width a, the directions of the minima in the diffraction pattern are given by:
A. $a\sin\theta = n\lambda/2$
B. $a\sin\theta = n\lambda$ (n = 1, 2, …) ✓ Correct
C. $a\cos\theta = n\lambda$
D. $a\sin\theta = (2n+1)\lambda/2$
Solution: Single-slit minima occur where a sinθ = nλ, n = 1, 2, 3, …
Q11 — Diffraction · easy · theory
In a single-slit diffraction pattern, the central maximum is:
A. Absent
B. The dimmest
C. The same width as the others
D. The brightest and the widest ✓ Correct
Solution: The central maximum carries most of the energy — it is the brightest and about twice as wide as the secondary maxima.
Q12 — Diffraction · easy · theory
Diffraction of light provides direct evidence for the _____ nature of light.
A. Particle
B. Wave ✓ Correct
C. Corpuscular
D. Quantum
Solution: Diffraction is a characteristic wave phenomenon, so it demonstrates the wave nature of light.
Q13 — Diffraction · easy · theory
In a single-slit diffraction pattern, the point directly opposite the centre of the slit is:
A. The second minimum
B. The first secondary maximum
C. A dark fringe
D. The centre of the bright central maximum ✓ Correct
Solution: At θ = 0 all secondary wavelets arrive in phase, giving the central bright maximum.
Q14 — Interference · hard · numerical
A thin sheet (μ = 1.5, thickness 10 μm) covers one slit. The number of fringes the pattern shifts (λ = 500 nm) is:
A. 10 ✓ Correct
B. 15
C. 5
D. 20
Solution: Shift (in fringes) = (μ − 1)t/λ = 0.5 × 10 × 10⁻⁶/500 × 10⁻⁹ = 10.
Q15 — Interference · hard · numerical
Light of wavelengths 6000 Å and 4800 Å illuminate a double slit. The 4th bright fringe of 6000 Å coincides with the nth bright fringe of 4800 Å, where n =
A. 3
B. 4
C. 5 ✓ Correct
D. 6
Solution: n₁λ₁ = n₂λ₂ ⇒ 4 × 6000 = n × 4800 ⇒ n = 5.
Q16 — Interference · hard · numerical
For two equal-intensity beams (I_max = 4I₀), the intensity where the path difference is λ/3 is:
A. 3I₀
B. I₀ ✓ Correct
C. 2I₀
D. 4I₀
Solution: φ = 2π/3; I = 4I₀cos²(π/3) = 4I₀(1/4) = I₀.
Q17 — Interference · hard · numerical
A mica sheet (μ = 1.6, thickness 2 μm) is placed over one slit (λ = 500 nm). The fringe shift is:
A. 2 fringes
B. 1.2 fringes
C. 2.4 fringes ✓ Correct
D. 4 fringes
Solution: Shift = (μ − 1)t/λ = 0.6 × 2 × 10⁻⁶/500 × 10⁻⁹ = 2.4 fringes.
Q18 — Diffraction · hard · numerical
The Fresnel distance for a beam of width 4 mm and wavelength 400 nm is:
A. 40 m ✓ Correct
B. 400 m
C. 4 m
D. 0.4 m
Solution: z_F = a²/λ = (4 × 10⁻³)²/(400 × 10⁻⁹) = 16 × 10⁻⁶/4 × 10⁻⁷ = 40 m.
Q19 — Diffraction · hard · numerical
The Fresnel distance for an aperture of 3 mm and light of 500 nm is:
A. 1.8 m
B. 0.18 m
C. 18 m ✓ Correct
D. 180 m
Solution: z_F = a²/λ = (3 × 10⁻³)²/(500 × 10⁻⁹) = 9 × 10⁻⁶/5 × 10⁻⁷ = 18 m.
Q20 — Diffraction · hard · numerical
The position of the 2nd minimum of a single-slit pattern (a = 0.2 mm, λ = 600 nm, D = 1 m) is:
A. 3 mm
B. 6 mm ✓ Correct
C. 12 mm
D. 1.5 mm
Solution: y = nλD/a = 2 × 600 × 10⁻⁹ × 1/0.2 × 10⁻³ = 6 mm.
Q21 — Diffraction · hard · numerical
According to the Rayleigh criterion, the limit of resolution of a telescope of aperture 0.1 m for 500 nm light is:
A. 5 × 10⁻⁶ rad
B. 6.1 × 10⁻⁶ rad ✓ Correct
C. 6.1 × 10⁻⁵ rad
D. 1.22 × 10⁻⁶ rad
Solution: θ = 1.22λ/D = 1.22 × 500 × 10⁻⁹/0.1 = 6.1 × 10⁻⁶ rad.
Q22 — Diffraction · hard · numerical
A single slit produces a central maximum of angular width 2 × 10⁻³ rad with 500 nm light. The slit width is:
A. 1 mm
B. 0.1 mm
C. 0.25 mm
D. 0.5 mm ✓ Correct
Solution: 2λ/a = 2 × 10⁻³ ⇒ a = 2 × 500 × 10⁻⁹/2 × 10⁻³ = 0.5 mm.
Q23 — Diffraction · hard · numerical
A slit of width 0.6 mm gives a central maximum of linear width 4 mm on a screen 2 m away. The wavelength of the light is:
A. 600 nm ✓ Correct
B. 400 nm
C. 500 nm
D. 750 nm
Solution: Width = 2λD/a ⇒ λ = (width × a)/(2D) = (4 × 10⁻³ × 0.6 × 10⁻³)/(2 × 2) = 6 × 10⁻⁷ m = 600 nm.
Q24 — Interference · medium · theory
The phase difference corresponding to constructive interference is:
A. $(2n+1)\pi/2$
B. $n\pi/2$
C. $(2n-1)\pi$
D. $2n\pi$ ✓ Correct
Solution: A path difference of nλ corresponds to a phase difference of 2nπ, giving constructive interference.
Q25 — Interference · medium · theory
Two coherent waves each of amplitude a and intensity I₀ interfere constructively. The resultant intensity is:
A. $4I_0$ ✓ Correct
B. Zero
C. $I_0$
D. $2I_0$
Solution: Amplitudes add: A = 2a, and intensity ∝ A², so I = (2a)² ∝ 4 × (a²) = 4I₀.
Q26 — Interference · medium · theory
The resultant intensity of two coherent waves of intensities I₁ and I₂ with phase difference φ is:
A. $\sqrt{I_1 I_2}\cos\phi$
B. $I_1 + I_2$
C. $I_1 + I_2 - 2\sqrt{I_1 I_2}\cos\phi$
D. $I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ ✓ Correct
Solution: The interference intensity is I = I₁ + I₂ + 2√(I₁I₂)cosφ.
Q27 — Interference · medium · theory
For two interfering beams of intensities I₁ and I₂, the maximum and minimum intensities are:
A. $(\sqrt{I_1}+\sqrt{I_2})^2$ and $(\sqrt{I_1}-\sqrt{I_2})^2$ ✓ Correct
B. $(I_1+I_2)^2$ and $(I_1-I_2)^2$
C. $(I_1+I_2)$ and $(I_1-I_2)$
D. $2I_1$ and $2I_2$
Solution: I_max = (√I₁ + √I₂)² (φ = 0) and I_min = (√I₁ − √I₂)² (φ = π).
Q28 — Interference · medium · numerical
In an interference pattern, the amplitudes of the two waves are in the ratio 3 : 1. The ratio of maximum to minimum intensity is:
A. 16 : 1
B. 4 : 1 ✓ Correct
C. 9 : 1
D. 3 : 1
Solution: I_max/I_min = ((a₁+a₂)/(a₁−a₂))² = ((3+1)/(3−1))² = (4/2)² = 4, i.e. 4 : 1.
Q29 — Interference · medium · numerical
Two coherent sources have intensities in the ratio 4 : 1. The ratio of maximum to minimum intensity in the interference pattern is:
A. 16 : 1
B. 4 : 1
C. 5 : 3
D. 9 : 1 ✓ Correct
Solution: Amplitude ratio = √4 : √1 = 2 : 1. I_max/I_min = ((2+1)/(2−1))² = 9 : 1.
Q30 — Interference · medium · theory
In Young's experiment, if the distance D between the slits and the screen is increased, the fringe width:
A. Decreases
B. Becomes zero
C. Stays the same
D. Increases ✓ Correct
Solution: β = λD/d, so increasing D increases the fringe width (fringes spread out).