Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be _________ I.

Solution

Physics sir · 4 steps

Add amplitudes, not intensities: take square roots, combine them, then square back.

1. Intensity formula. Two beams of intensities I₁ and I₂ with phase difference φ give a resultant intensity that depends on cos φ.

   I = I₁ + I₂ + 2√(I₁ I₂)cos φ

2. Point A. Here φ = 0, so cos φ = 1. With I₁ = 4I and I₂ = 9I, √(I₁ I₂) = 6I, so I_A = 4I + 9I + 12I = 25I. This is the same as (√4 + √9)² I.
3. Point B. Here φ = π, so cos φ = -1. That gives I_B = 4I + 9I - 12I = I, the same as (√9 - √4)² I.
4. So. I_A - I_B = 25I - I = 24I, so the answer is 24.

The common slip: 12. Taking the cross term 2√(I₁ I₂) only once gives 12, not 24. It is added at A and subtracted at B, so the difference is twice that, 4√(I₁ I₂).

The idea to keep. Bright and dark intensities are (√I₁ ± √I₂)², so their difference is always 4√(I₁ I₂). That gives the answer in one line.

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