Wave Optics — Class 12 MCQs with Answers

Wave Optics — important questions & MCQs with answers (Class 12 Physics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Young's experiment demonstrates:

   A.Interference of light✓ Correct

   B.Reflection only

   C.Refraction only

   D.Photoelectric effect

   Answer: A. Interference of light

   Explanation: Young's double-slit setup overlaps light from two coherent slits, and the alternating bright/dark fringes it produces are a wave signature — only interference explains that pattern, not reflection or refraction.

2. Q2Easy

   Light is polarised because it is:

   A.Sound

   B.Longitudinal

   C.Mechanical

   D.A transverse wave✓ Correct

   Answer: D. A transverse wave

   Explanation: A wave can be polarised only if its vibrations are perpendicular to its direction of travel, i.e. transverse; sound is longitudinal, with no transverse direction to restrict, so it cannot be polarised.

3. Q3Easy

   A wavefront is the locus of points having:

   A.The same amplitude only

   B.The same phase✓ Correct

   C.The same frequency only

   D.Zero phase

   Answer: B. The same phase

   Explanation: A wavefront is defined as the locus of points vibrating in phase, not equal amplitude — points on the same wavefront can have different amplitudes, but their phase is always identical.

4. Q4Easy

   Diffraction is:

   A.Total reflection

   B.Splitting of white light

   C.Bending of light around obstacles✓ Correct

   D.Polarisation

   Answer: C. Bending of light around obstacles

   Explanation: Diffraction is the bending of light into the geometric shadow around obstacles or through narrow slits — a direct consequence of light's wave nature, distinct from reflection or refraction.

5. Q5Easy

   A point source produces a wavefront that is:

   A.Linear

   B.Plane

   C.Cylindrical

   D.Spherical✓ Correct

   Answer: D. Spherical

   Explanation: A point source emits equally in all directions, so surfaces of constant phase form concentric spheres centred on it — a spherical wavefront, which looks plane only far away.

6. Q6Easy

   Malus's law applies to:

   A.Diffraction maxima

   B.Intensity of polarised light through analyser✓ Correct

   C.Newton's rings

   D.Refraction

   Answer: B. Intensity of polarised light through analyser

   Explanation: Malus's law, I = I₀cos²θ, gives the intensity of already-polarised light passing through an analyser tilted at angle θ to the polarisation axis — not diffraction or refraction.

7. Q7Medium

   Fringe width β =

   A.λ/d

   B.d/λD

   C.λD/d✓ Correct

   D.λD

   Answer: C. λD/d

   Explanation: Fringe width in YDSE is β = λD/d — proportional to wavelength and screen distance D, inversely proportional to slit separation d; swapping D and d in the formula is a common error.

8. Q8Medium

   Width of central maximum in single-slit diffraction (screen at D, slit a):

   A.λ/a

   B.λD/a

   C.2λD/a✓ Correct

   D.a/D

   Answer: C. 2λD/a

   Explanation: The full width of the central maximum in single-slit diffraction, at screen distance D, is 2λD/a — twice the distance λD/a to a single first minimum.

9. Q9Medium

   For sustained interference, sources must be:

   A.Incoherent

   B.Coherent (constant phase)✓ Correct

   C.Different colours

   D.Random

   Answer: B. Coherent (constant phase)

   Explanation: Sustained, stable interference fringes need coherent sources — same frequency and a constant phase difference — so the pattern doesn't drift and blur from randomly varying phase.

10. Q10Medium

    Unpolarised light I₀ through two polaroids at 60°. Final intensity:

    A.I₀/8✓ Correct

    B.I₀/4

    C.I₀/2

    D.I₀

    Answer: A. I₀/8

    Explanation: Unpolarised I₀ first halves through the polariser to I₀/2, then Malus's law at 60° gives (I₀/2)cos²60° = (I₀/2)(¼) = I₀/8 — forgetting the initial halving is the usual slip.

11. Q11Medium

    λ = 600 nm, d = 0.6 mm, D = 1 m. Fringe width:

    A.0.6 mm

    B.0.1 mm

    C.6 mm

    D.1 mm✓ Correct

    Answer: D. 1 mm

    Explanation: β = λD/d = (600×10⁻⁹ × 1)/(0.6×10⁻³) = 10⁻³ m = 1 mm — convert every length to metres before dividing, to avoid a stray factor of 10.

12. Q12Hard

    YDSE: d=1mm, D=1m, λ=600nm. A slab t=10μm, n=1.5 covers one slit. Fringe shift:

    A.0.5 mm

    B.10 mm

    C.5 mm✓ Correct

    D.1 mm

    Answer: C. 5 mm

    Explanation: Fringe shift Δ = (n−1)tD/d = (0.5)(10×10⁻⁶)(1)/(10⁻³) = 5×10⁻³ m = 5 mm — it's (n−1), the extra path beyond what air alone would give, not n itself.

13. Q13Hard

    Single-slit diffraction first minimum:

    A.a sinθ = nλ/2

    B.a sinθ = λ✓ Correct

    C.a = λ

    D.sinθ = a

    Answer: B. a sinθ = λ

    Explanation: The general single-slit minima condition a sinθ = nλ gives the first minimum at n=1: a sinθ = λ — the nλ/2 form belongs to a different, two-source interference condition.

14. Q14Hard

    If slit width a and separation d satisfy d = 3a, missing orders are:

    A.All

    B.n = 2, 4

    C.None

    D.n = 3, 6, 9, …✓ Correct

    Answer: D. n = 3, 6, 9, …

    Explanation: Orders vanish where an interference maximum coincides with a diffraction minimum: n·a = m·d ⇒ n = 3m, so the missing orders are n = 3, 6, 9, … — every third order disappears since d/a=3.

Wave Optics — FAQs

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