In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _________ × 10⁻⁶ m.

Solution

Physics sir · 4 steps

Half intensity means a phase difference of π/2, so you are just a quarter of a fringe from the centre.

1. Intensity and phase. For two equal slits, I = I₀ cos²(φ/2), where φ is the phase difference. Setting I = I₀/2 gives cos²(φ/2) = 1/2, so the first time this happens is at φ/2 = π/4, that is φ = π/2.
2. Path difference. A phase difference of π/2 is a path difference of λ/2π × π/2 = λ/4. On the screen the path difference is yd/D.

   yd/D = λ/4 ⇒ y = (λ D)/4d

3. Put in numbers. With λ = 5000 Å = 5 × 10⁻⁷ m, D = 1.0 m and d = 1.0 × 10⁻³ m: y = (5 × 10⁻⁷ × 1)/(4 × 10⁻³) = 1.25 × 10⁻⁴ m.
4. So. y = 125 × 10⁻⁶ m, a quarter of the fringe width β = 500 × 10⁻⁶ m. The answer is 125.

The common slip: 250. Taking the path difference as λ/2 instead of λ/4 gives 250, which is where the intensity drops to zero, not to half.

The idea to keep. In YDSE, read intensity as a phase: I₀/2 needs phase π/2, which is path λ/4, which is β/4 on the screen. Zero intensity is β/2.

Answer from the NTA final answer key. Solution by PrepWiser.