Wave Phenomena and Light

Fringe width decreases by a factor of 5/6

Step 1: Fringe width β = λD/d. Since D and d remain unchanged, β is directly proportional to λ. Step 2: β_yellow = 600D/d and β_green = 500D/d. Step 3: Ratio β_green/β_yellow = 500/600 = 5/6. Step 4: Since 5/6 < 1, the new fringe width is SMALLER than the original. So fringe width decreases by a factor of 5/6. Step 5: Option A is wrong — it incorrectly states an increase. Option C is wrong because fringe width clearly depends on λ. Option D confuses 'decrease' with 'increase'.

sin θ = λ/a

Step 1: In single slit diffraction, minima (dark fringes) occur when the path difference between waves from the two extreme edges of the slit is an integer multiple of λ. Step 2: The condition for the nth minimum is: a sin θ = nλ. Step 3: For the FIRST minimum (n = 1): a sin θ = λ → sin θ = λ/a. Step 4: The central maximum extends from −θ to +θ, giving it twice the angular width compared to secondary maxima (which span from one minimum to the next). Step 5: Option A (λ/2a) would be half the correct value. Option C (2λ/a) corresponds to the second minimum, not the first.

56.3°

Step 1: Brewster's Law states: tan(iₚ) = μ, where iₚ is the polarising angle and μ is the refractive index. Step 2: Given μ = 1.5, so tan(iₚ) = 1.5. Step 3: iₚ = tan⁻¹(1.5) ≈ 56.3°. Step 4: At Brewster's angle, the reflected ray is completely plane polarised. Also, the reflected ray and the refracted ray are perpendicular to each other (iₚ + r = 90°). Step 5: Option A (33.7°) is actually the refraction angle r = 90° − 56.3°. Option C (45°) corresponds to μ = 1, which is not glass. Option D (63.4°) corresponds to μ ≈ 2.0.

4I₀

Step 1: Intensity of a single wave: I₀ ∝ a². Step 2: For constructive interference (phase difference δ = 2nπ), the resultant amplitude A = 2a. Step 3: Intensity at bright fringe: I_max ∝ A² = (2a)² = 4a². Step 4: Therefore, I_max = 4a² = 4I₀ (since I₀ ∝ a²). Step 5: Note: Energy is conserved — the average intensity over the pattern remains 2I₀ (as dark fringes have zero intensity). Option B (2I₀) would be the average, not the maximum. Option D (√2 I₀) has no physical basis here.