In a Young's double slits experiment, the ratio of amplitude of light coming from slits is 2: 1. The ratio of the maximum to minimum intensity in the interference pattern is:

Solution

Physics sir · 3 steps

Amplitudes add at a bright fringe and subtract at a dark one. Square both at the end to get intensity.

1. Take the amplitudes. Let the two amplitudes be a₁ = 2a and a₂ = a. Intensity is proportional to the square of the resultant amplitude.
2. Bright and dark. At a maximum the waves meet in phase, so the resultant amplitude is a₁ + a₂ = 3a. At a minimum they meet out of phase, so it is a₁ - a₂ = a.

   Iₘₐₓ/Iₘᵢₙ = (a₁ + a₂)²/(a₁ - a₂)² = (3a)²/(a)² = 9/1

3. So. The ratio of maximum to minimum intensity is 9 : 1, which is option (b).

The idea to keep. Never add intensities in interference. Add or subtract the amplitudes first, then square. If you are given an intensity ratio instead, take square roots to get amplitudes before you start.

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