In a Young's double slit experiment, the intensity at a point is (1/4)ᵗʰ of the maximum intensity, the minimum distance of the point from the central maximum is _________ μ m. (Given : λ=600 nm, d=1.0 mm, D=1.0 m)

Solution

Physics sir · 4 steps

Turn the intensity ratio into a phase difference, then the phase into a path difference and a distance.

1. Intensity to phase. For two equal slits, I = Iₘₐₓ cos²(φ/2). Setting I = Iₘₐₓ/4 gives cos(φ/2) = 1/2. The smallest solution is φ/2 = π/3, so φ = 2π/3.
2. Phase to path. Path difference is Δ x = λ/2πφ = λ/3. With λ = 600 nm, that is Δ x = 200 nm.

   y = (Δ x D)/d = (200 × 10⁻⁹)(1.0)/(1.0 × 10⁻³) = 2 × 10⁻⁴ m

3. Quick check. The fringe width is β = (λ D)/d = 600 μm. A phase of 2π/3 is one third of a full fringe, so y = β/3 = 200 μm.
4. So. The point is 2 × 10⁻⁴ m from the central maximum, which is 200 μm.

The common slip: 100. Dropping the half in cos²(φ/2) and taking φ = π/3 gives 100, not 200.

The idea to keep. In YDSE, I = Iₘₐₓ cos²(φ/2). Get the phase from the intensity ratio, then y = φ/2πβ, with β = (λ D)/d.

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