In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7 λ / 4. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

Solution

Physics sir · 4 steps

Turn the path difference into a phase difference, then halve it and square the cosine.

1. Phase difference. A path difference Δ x gives a phase difference φ = 2π/λΔ x. Here φ = 2π/λ × 7λ/4 = 7π/2.
2. Intensity formula. For two identical sources, each of intensity I₀, the resultant intensity is 4I₀cos²(φ/2). The maximum is 4I₀, so the ratio below follows.

   I/Iₘₐₓ = cos² φ/2 = cos² 7π/4

3. Evaluate. 7π/4 lies in the fourth quadrant, where cos 7π/4 = cos π/4 = 1/√2. Squaring gives 1/2.
4. So. The intensity at this point is half the maximum intensity, option (a).

The idea to keep. For identical sources, I/Iₘₐₓ = cos²(φ/2) with φ = 2πΔ x/λ. Whole wavelengths in Δ x change nothing, so only the leftover fraction of λ matters.

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