The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be :

Solution

Physics sir · 4 steps

Same two beams at both points, so only the cosine term changes: find it at P and at Q.

1. Interference formula. When two beams of intensities I₁ and I₂ meet with phase difference φ, the resultant intensity is given below. Here √(I₁ I₂) = √(I × 9I) = 3I.

   Iᵣₑₛ = I₁ + I₂ + 2√(I₁ I₂)cos φ = 10I + 6I cos φ

2. At point P. φ = π/2, so cos φ = 0. The cross term vanishes and I_P = 10I.
3. At point Q. φ = π, so cos φ = -1. Then I_Q = 10I - 6I = 4I, which is the minimum intensity (√9I - √I)².
4. So. I_P - I_Q = 10I - 4I = 6I, which is option (b).

The idea to keep. For two beams, I = I₁ + I₂ + 2√(I₁ I₂)cos φ. At φ = π/2 the intensities just add; at φ = π you get the minimum (√I₁ - √I₂)².

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