Interference — IISER Physics MCQs with Solutions
Free IISER Physics Interference MCQs with step-by-step solutions (54 questions). Part of Interference and Diffraction of Light. Practise online on Prepizo — no login needed.
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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 · medium · theory
The phase difference corresponding to constructive interference is:
A. $n\pi/2$
B. $(2n-1)\pi$
C. $2n\pi$ ✓ Correct
D. $(2n+1)\pi/2$
Solution: A path difference of nλ corresponds to a phase difference of 2nπ, giving constructive interference.
Q6 — Interference · medium · theory
Two coherent waves each of amplitude a and intensity I₀ interfere constructively. The resultant intensity is:
A. $2I_0$
B. $I_0$
C. $4I_0$ ✓ Correct
D. Zero
Solution: Amplitudes add: A = 2a, and intensity ∝ A², so I = (2a)² ∝ 4 × (a²) = 4I₀.
Q7 — Interference · medium · theory
The resultant intensity of two coherent waves of intensities I₁ and I₂ with phase difference φ is:
A. $I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ ✓ Correct
B. $I_1 + I_2$
C. $I_1 + I_2 - 2\sqrt{I_1 I_2}\cos\phi$
D. $\sqrt{I_1 I_2}\cos\phi$
Solution: The interference intensity is I = I₁ + I₂ + 2√(I₁I₂)cosφ.
Q8 — Interference · medium · theory
For two interfering beams of intensities I₁ and I₂, the maximum and minimum intensities are:
A. $(I_1+I_2)$ and $(I_1-I_2)$
B. $(I_1+I_2)^2$ and $(I_1-I_2)^2$
C. $(\sqrt{I_1}+\sqrt{I_2})^2$ and $(\sqrt{I_1}-\sqrt{I_2})^2$ ✓ Correct
D. $2I_1$ and $2I_2$
Solution: I_max = (√I₁ + √I₂)² (φ = 0) and I_min = (√I₁ − √I₂)² (φ = π).
Q9 — 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. 4 : 1 ✓ Correct
B. 3 : 1
C. 16 : 1
D. 9 : 1
Solution: I_max/I_min = ((a₁+a₂)/(a₁−a₂))² = ((3+1)/(3−1))² = (4/2)² = 4, i.e. 4 : 1.
Q10 — 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. 5 : 3
B. 16 : 1
C. 4 : 1
D. 9 : 1 ✓ Correct
Solution: Amplitude ratio = √4 : √1 = 2 : 1. I_max/I_min = ((2+1)/(2−1))² = 9 : 1.
Q11 — 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.
Q12 — Interference · medium · theory
In Young's experiment, if the distance D between the slits and the screen is increased, the fringe width:
A. Stays the same
B. Becomes zero
C. Decreases
D. Increases ✓ Correct
Solution: β = λD/d, so increasing D increases the fringe width (fringes spread out).
Q13 — Interference · medium · theory
If the separation d between the two slits is increased, the fringe width:
A. Becomes infinite
B. Increases
C. Decreases ✓ Correct
D. Stays the same
Solution: β = λD/d is inversely proportional to d, so increasing d decreases the fringe width.
Q14 — Interference · medium · theory
In Young's double-slit experiment, which colour produces the widest fringes?
A. Green
B. Blue
C. Red ✓ Correct
D. Violet
Solution: β ∝ λ; red light has the longest wavelength, so it gives the widest fringes.
Q15 — Interference · medium · numerical
In a Young's experiment, λ = 600 nm, slit separation d = 1 mm, screen distance D = 1 m. The fringe width is:
A. 0.6 mm ✓ Correct
B. 1.2 mm
C. 6 mm
D. 0.06 mm
Solution: β = λD/d = (600 × 10⁻⁹ × 1)/(1 × 10⁻³) = 6 × 10⁻⁴ m = 0.6 mm.
Q16 — Interference · medium · theory
The distance of the nth bright fringe from the central maximum in Young's experiment is:
A. $y_n = \dfrac{n\lambda d}{D}$
B. $y_n = \dfrac{\lambda D}{nd}$
C. $y_n = \dfrac{(2n-1)\lambda D}{2d}$
D. $y_n = \dfrac{n\lambda D}{d}$ ✓ Correct
Solution: The nth bright fringe lies at y_n = nλD/d = nβ.
Q17 — Interference · medium · theory
The distance of the nth dark fringe from the central maximum is:
A. $\dfrac{(2n+1)\lambda D}{d}$
B. $\dfrac{(2n-1)\lambda D}{2d}$ ✓ Correct
C. $\dfrac{n\lambda D}{2d}$
D. $\dfrac{n\lambda D}{d}$
Solution: Dark fringes occur where the path difference is (2n−1)λ/2, giving y = (2n−1)λD/2d.
Q18 — Interference · medium · theory
In Young's double-slit experiment, the fringe width is:
A. Smaller for higher-order fringes
B. Zero for even orders
C. Larger for higher-order fringes
D. The same for all fringes (independent of order) ✓ Correct
Solution: All bright (and dark) fringes are equally spaced; the fringe width β = λD/d does not depend on the order n.
Q19 — Interference · medium · theory
When the whole Young's double-slit apparatus is immersed in water (μ = 1.33), the fringe width:
A. Decreases (becomes β/1.33) ✓ Correct
B. Remains unchanged
C. Increases (becomes 1.33β)
D. Becomes zero
Solution: The wavelength in water is λ/μ, so β' = λD/(μd) = β/μ — the fringes get narrower.
Q20 — Interference · medium · theory
When white light is used in Young's double-slit experiment, the central fringe is:
A. Red
B. White ✓ Correct
C. Dark
D. Violet
Solution: At the centre the path difference is zero for all wavelengths, so all colours reinforce and the central fringe is white (the others are coloured).
Q21 — Interference · medium · theory
If one of the two slits in Young's experiment is completely covered, the pattern on the screen shows:
A. More fringes
B. Brighter fringes
C. Almost uniform illumination (no interference fringes) ✓ Correct
D. The same fringes
Solution: With only one slit open there is no second coherent beam to interfere with, so the fringes disappear and the screen is nearly uniformly illuminated.
Q22 — Interference · medium · theory
Two independent light bulbs cannot produce a sustained interference pattern because:
A. They do not maintain a constant phase difference (they are incoherent) ✓ Correct
B. They have different colours
C. They emit polarised light
D. They are too bright
Solution: Independent sources emit light in random, rapidly changing phases, so the phase difference is not constant and no steady interference pattern forms.
Q23 — Interference · medium · theory
The angular fringe width in Young's double-slit experiment is:
A. $\dfrac{\lambda D}{d}$
B. $\dfrac{d}{\lambda}$
C. $\dfrac{\lambda}{D}$
D. $\dfrac{\lambda}{d}$ ✓ Correct
Solution: Angular fringe width = β/D = λ/d.
Q24 — Interference · medium · numerical
A Young's double-slit pattern has a fringe width of 0.5 mm. If the screen distance D is doubled (everything else unchanged), the new fringe width is:
A. 0.5 mm
B. 1.0 mm ✓ Correct
C. 2.0 mm
D. 0.25 mm
Solution: β ∝ D, so doubling D doubles the fringe width: 0.5 mm → 1.0 mm.
Q25 — Interference · medium · numerical
In a Young's experiment, the fringe width is 0.6 mm when d = 1 mm and D = 1 m. The wavelength of light used is:
A. 500 nm
B. 400 nm
C. 750 nm
D. 600 nm ✓ Correct
Solution: λ = βd/D = (0.6 × 10⁻³ × 1 × 10⁻³)/1 = 6 × 10⁻⁷ m = 600 nm.
Q26 — Interference · medium · theory
A thin transparent sheet of thickness t and refractive index μ is placed in front of one slit. The fringe pattern shifts:
A. Towards the side of the slit with the sheet, by (μ − 1)tD/d ✓ Correct
B. Away from the sheet
C. By (μ + 1)tD/d away from the sheet
D. Without any shift
Solution: The sheet adds an extra optical path (μ − 1)t to that slit, shifting the whole pattern towards that slit by (μ − 1)tD/d.
Q27 — Interference · medium · theory
The number of fringes by which the pattern shifts when a sheet of thickness t and refractive index μ covers one slit is:
A. $\dfrac{(\mu-1)\lambda}{t}$
B. $\dfrac{\mu t}{\lambda}$
C. $\dfrac{(\mu-1)t}{2\lambda}$
D. $\dfrac{(\mu-1)t}{\lambda}$ ✓ Correct
Solution: Fringe shift in number of fringes = extra path / λ = (μ − 1)t / λ.
Q28 — Interference · medium · theory
If the slit separation d is halved in Young's experiment (D and λ unchanged), the fringe width:
A. Becomes one quarter
B. Halves
C. Doubles ✓ Correct
D. Is unchanged
Solution: β = λD/d is inversely proportional to d, so halving d doubles the fringe width.
Q29 — Interference · medium · theory
Two wavelengths λ₁ and λ₂ are used together in Young's experiment. The nth bright fringe of λ₁ coincides with the mth bright fringe of λ₂ when:
A. $\lambda_1 + \lambda_2 = nm$
B. $n\lambda_1 = m\lambda_2/2$
C. $n\lambda_2 = m\lambda_1$
D. $n\lambda_1 = m\lambda_2$ ✓ Correct
Solution: Bright fringes coincide when their positions are equal: nλ₁D/d = mλ₂D/d ⇒ nλ₁ = mλ₂.
Q30 — Interference · medium · theory
In a two-slit interference pattern with equal-intensity beams (each I₀ giving maximum 4I₀), the intensity at a point where the path difference is λ/4 is:
A. $4I_0$
B. $I_0$
C. Zero
D. $2I_0$ ✓ Correct
Solution: Path difference λ/4 ⇒ φ = π/2. I = 4I₀cos²(φ/2) = 4I₀cos²(π/4) = 4I₀ × 1/2 = 2I₀.