Interference of Light — Important Questions

0.9 mm

Step 1: Fringe width β = Dλ/d. When the apparatus is immersed in a medium of refractive index n, the wavelength changes: λ_medium = λ_air / n. Step 2: New fringe width β' = Dλ_medium/d = β/n. Step 3: β' = 1.2 mm / (4/3) = 1.2 × (3/4) = 0.9 mm. Option A (1.6 mm) is incorrect – immersing in water decreases the wavelength, so fringe width decreases, not increases. Option D (0.6 mm) would correspond to n = 2.

The two lamps are incoherent – their phase difference changes randomly and continuously.

Step 1: For sustained interference, two sources must be coherent – they must maintain a constant phase difference over time. Step 2: Independent sources consist of billions of atoms emitting light randomly; each atom emits for only ~10⁻⁸ s and then stops, so the phase relationship between two lamps changes billions of times per second. Step 3: Because the phase difference fluctuates randomly, the positions of bright and dark fringes shift so rapidly that the eye (due to persistence of vision) perceives uniform illumination – no fringes are visible. Option A is incorrect because sodium lamps em

600 nm

Step 1: Use x_m = mDλ/d, so λ = x_m × d / (m × D). Step 2: Given: x₄ = 1.2 cm = 1.2 × 10⁻² m, d = 0.28 mm = 0.28 × 10⁻³ m, D = 1.4 m, m = 4. Step 3: λ = (1.2 × 10⁻² × 0.28 × 10⁻³) / (4 × 1.4) = (3.36 × 10⁻⁶) / 5.6 = 6 × 10⁻⁷ m. Step 4: λ = 600 nm. This is a standard NCERT problem. Options A, B, D result from arithmetic errors in the calculation.

Δβ = λ·ΔD / d

Step 1: Fringe width β = Dλ/d. Step 2: Since λ and d are constant, any change in D directly causes a proportional change in β: Δβ = λ·ΔD / d. Step 3: This follows because β is directly proportional to D. Step 4: This result is used in ISC problems where the screen is moved and the change in fringe width is measured to find λ. Options B, C, D have incorrect arrangements of variables and do not follow from the formula.