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Interference and Diffraction of Light — IISER 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. Superposition of waves ✓ Correct
B. Rectilinear propagation
C. Total internal reflection
D. Quantisation of energy
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. The same frequency and a constant phase difference ✓ Correct
B. A randomly varying phase difference
C. Different frequencies
D. The same amplitude only
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. $n\lambda/2$
D. $(2n-1)\lambda/2$
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. $n\lambda/4$
B. $n\lambda$
C. $(2n-1)\dfrac{\lambda}{2}$ ✓ Correct
D. $2n\lambda$
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. A bright fringe (zero path difference) ✓ Correct
C. Uniformly grey
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. π
B. 2π ✓ Correct
C. 4π
D. π/2
Solution: Phase difference = (2π/λ) × path difference = (2π/λ) × λ = 2π.
Q8 — Diffraction · easy · theory
Diffraction of light is the phenomenon of:
A. Splitting of white light into colours
B. Reflection at a polished surface
C. Rotation of the plane of vibration
D. Bending of light around the edges of an obstacle or aperture ✓ Correct
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. Exactly one metre
B. Comparable to the wavelength of light ✓ Correct
C. Very much larger than the wavelength
D. Independent of the wavelength
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. $a\sin\theta = (2n+1)\lambda/2$
B. $a\sin\theta = n\lambda/2$
C. $a\sin\theta = n\lambda$ (n = 1, 2, …) ✓ Correct
D. $a\cos\theta = n\lambda$
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. The dimmest
B. The brightest and the widest ✓ Correct
C. The same width as the others
D. Absent
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. Corpuscular
C. Wave ✓ Correct
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. A dark fringe
B. The second minimum
C. The centre of the bright central maximum ✓ Correct
D. The first secondary maximum
Solution: At θ = 0 all secondary wavelets arrive in phase, giving the central bright maximum.
Q14 — Interference · hard · numerical
In YDSE the intensity at the central maximum is I₀. The intensity at a point where the path difference is λ/6 is:
A. 3I₀/4 ✓ Correct
B. I₀/√2
C. I₀/4
D. I₀/2
Solution: φ = 2π/6 = 60°; I = I₀cos²(φ/2) = I₀cos²30° = 3I₀/4.
Q15 — Interference · hard · numerical
In YDSE, the intensity at a point with path difference λ/8, as a fraction of the central maximum I₀, is (cos45° = 1/√2):
A. 0.25 I₀
B. 0.5 I₀
C. 0.71 I₀
D. ≈ 0.85 I₀ ✓ Correct
Solution: φ = 2π/8 = 45°; I = I₀cos²(22.5°) ≈ I₀ × 0.854.
Q16 — Interference · hard · numerical
Two slits have widths in the ratio 1 : 4 (intensity ∝ width). The ratio I_max : I_min in the pattern is:
A. 9 : 1 ✓ Correct
B. 16 : 1
C. 25 : 9
D. 4 : 1
Solution: Amplitudes 1 : 2 ⇒ (3/1)² = 9 : 1.
Q17 — Interference · hard · numerical
In YDSE with λ = 500 nm, d = 1 mm, D = 2 m, the separation between the 3rd bright fringes on opposite sides of the centre is:
A. 12 mm
B. 6 mm ✓ Correct
C. 3 mm
D. 1.5 mm
Solution: β = λD/d = 1 mm; y₃ = 3 mm each side ⇒ separation 6 mm.
Q18 — Interference · hard · numerical
In a YDSE the 8th bright fringe (λ₁ = 500 nm) coincides with the 10th bright fringe of another wavelength λ₂. Then λ₂ =
A. 400 nm ✓ Correct
B. 350 nm
C. 450 nm
D. 625 nm
Solution: 8λ₁ = 10λ₂ ⇒ λ₂ = 8×500/10 = 400 nm.
Q19 — Interference · hard · numerical
When a YDSE apparatus moves from air into a liquid, the fringe width falls to 3/4 of its value. The refractive index of the liquid is:
A. 3/4
B. 9/16
C. 1.5
D. 4/3 ✓ Correct
Solution: β ∝ λ = λ₀/n ⇒ n = β_air/β_liquid = 4/3.
Q20 — Interference · hard · numerical
In YDSE, the intensity at the central maximum is I₀. At the point where the path difference between the two waves is λ/4, the intensity is:
A. I₀/4
B. Zero
C. I₀/2 ✓ Correct
D. 3I₀/4
Solution: φ = (2π/λ)(λ/4) = π/2; I = I₀cos²(φ/2) = I₀cos²(π/4) = I₀/2.
Q21 — Interference · hard · numerical
In YDSE the slit separation is doubled AND the screen distance is halved. The fringe width becomes:
A. Unchanged
B. Double
C. One quarter ✓ Correct
D. Half
Solution: β = λD/d → λ(D/2)/(2d) = β/4.
Q22 — Interference · hard · numerical
Two coherent sources of intensities I and 4I interfere. The intensities at the maxima and minima are:
A. 5I and 3I
B. 4I and I
C. 25I and 9I
D. 9I and I ✓ Correct
Solution: I_max = (√I + 2√I)² = 9I; I_min = (2√I − √I)² = I.
Q23 — Interference · hard · numerical
In YDSE the distance of the 4th DARK fringe from the centre (λ = 600 nm, d = 0.6 mm, D = 1 m) is:
A. 2.5 mm
B. 4.5 mm
C. 3.5 mm ✓ Correct
D. 4 mm
Solution: y = (2n−1)λD/2d with n = 4: 7×600×10⁻⁹×1/(2×0.6×10⁻³) = 3.5 mm.
Q24 — Interference · hard · numerical
A YDSE uses white light (400–700 nm) with d = 1 mm and D = 1 m. At a point 0.75 mm from the central fringe, the visible wavelength that is ABSENT (dark) is:
A. 500 nm ✓ Correct
B. 600 nm
C. 400 nm
D. 750 nm
Solution: Path difference Δ = yd/D = 750 nm. Dark when Δ = (2n−1)λ/2 ⇒ λ = 1500, 500, 300 nm — only 500 nm lies in the visible band, so green is missing there.
Q25 — Interference · hard · numerical
In YDSE with λ = 600 nm, the smallest distance from the central maximum where the intensity drops to 25% of the maximum corresponds to a phase difference of:
A. 45°
B. 60°
C. 90°
D. 120° (path difference λ/3) ✓ Correct
Solution: I/I₀ = cos²(φ/2) = 1/4 ⇒ φ/2 = 60° ⇒ φ = 120°, Δ = λ/3.
Q26 — Interference · hard · numerical
A thin glass sheet (μ = 1.5) of thickness 4 μm covers one slit (λ = 600 nm). The central fringe shifts by how many fringe widths?
A. 10/3 ≈ 3.3 ✓ Correct
B. 10
C. 2
D. 5
Solution: Shift = (μ−1)t/λ = 0.5×4×10⁻⁶/6×10⁻⁷ = 10/3 fringes.
Q27 — Interference · hard · numerical
In a YDSE, when one slit is covered by a sheet of μ = 1.6 and thickness t, the central fringe moves to the position formerly occupied by the 6th bright fringe (λ = 600 nm). Then t =
A. 6 μm ✓ Correct
B. 10 μm
C. 3.6 μm
D. 1 μm
Solution: (μ−1)t = 6λ ⇒ t = 6×600×10⁻⁹/0.6 = 6 μm.
Q28 — Interference · hard · numerical
Two coherent beams give I_max/I_min = 25. The ratio of their amplitudes is:
A. 4 : 1
B. 25 : 1
C. 3 : 2 ✓ Correct
D. 5 : 1
Solution: (a₁+a₂)/(a₁−a₂) = 5 ⇒ a₁/a₂ = 6/4 = 3/2.
Q29 — Interference · hard · numerical
In YDSE (λ = 500 nm, d = 0.5 mm), the angular position of the 2nd dark fringe is:
A. 1 × 10⁻³ rad
B. 1.5 × 10⁻³ rad ✓ Correct
C. 2 × 10⁻³ rad
D. 2.5 × 10⁻³ rad
Solution: θ = (2n−1)λ/2d = 3×500×10⁻⁹/(2×0.5×10⁻³) = 1.5×10⁻³ rad.
Q30 — Interference · hard · numerical
The two slits of a YDSE emit with a constant phase difference of π (in addition to path effects). The centre of the screen then shows:
A. A dark fringe ✓ Correct
B. Intensity I₀/2
C. A bright fringe
D. No pattern
Solution: The built-in π shift converts the central maximum into a minimum — the whole pattern shifts by half a fringe.