Wave Optics — Exam Practice

Answer Wavelength decreases in a denser medium (λ_water = λ_air/n), and since fringe width β = λD/d, the fringe width decreases proportionally — the fringes become more closely spaced.

Answer For two slits separated by d, and a screen at distance D (D ≫ d), the path difference at a point y from the centre is Δ = dy/D.

Bright fringes occur where Δ = nλ, giving y_n = nλD/d. Consecutive bright fringes (n and n+1) are separated by:
β = y_n+1 − y_n = λD/d.

Answer (a) Two waves arriving with path difference Δ interfere constructively when Δ = nλ (crest meets crest), and destructively when Δ = (n+½)λ (crest meets trough), for integer n.

(b) y_n = nλD/d = (4 × 600×10⁻⁹ × 1) / (1×10⁻³) = 2400×10⁻⁶ m = 2.4 mm from the central maximum.

Solution Both conditions — matching frequency and a fixed (not randomly drifting) phase relationship — are needed for a stable, observable interference pattern.

Solution The central maximum spans between the first minima on either side, making it exactly twice the angular width of any secondary maximum.

Solution I = I₀cos²θ, where θ is the angle between the incoming polarised light's direction and the analyser's transmission axis.

Solution Only transverse waves — which oscillate perpendicular to their direction of travel — can be restricted to a single plane of vibration. Longitudinal waves (like sound) cannot be polarised at all.

Solution β = λD/d = (500×10⁻⁹ × 1.5) / (0.5×10⁻³) = 750×10⁻⁹ / 0.0005 = 1.5×10⁻³ m = 1.5 mm.

Solution tanθ_B = n, and at this specific angle, the reflected and refracted rays are exactly perpendicular to each other.

Solution After the first polariser: I₁ = I₀/2 (unpolarised light is always halved by a single polariser).

After the second, by Malus's Law: I₂ = I₁cos²(45°) = (I₀/2)(0.5) = I₀/4.

Fraction transmitted = 1/4 (25%).