Two coherent monochromatic light beams of intensities I and 4 I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is x I. The value of x is __________.

Solution

Physics sir · 4 steps

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

1. Amplitudes first. Intensity goes as amplitude squared, so amplitude goes as √I. The two amplitudes are in the ratio √I : √4I = 1 : 2.

   Iₘₐₓ = (√I₁+√I₂)², Iₘᵢₙ = (√I₁-√I₂)²

2. Put in numbers. Iₘₐₓ = (√I + 2√I)² = 9I. Iₘᵢₙ = (2√I - √I)² = I.
3. Shortcut check. The difference is always Iₘₐₓ - Iₘᵢₙ = 4√(I₁ I₂) = 4√4I² = 8I, which matches.
4. So. Iₘₐₓ - Iₘᵢₙ = 9I - I = 8I, so x = 8.

The common slip: 16. Adding the intensities as if they were amplitudes, (1+4)² - (4-1)², gives 16, not 8. Take square roots of the intensities first.

The idea to keep. In interference, amplitudes add and subtract, never intensities. Take √I of each beam, combine, then square. The gap between maximum and minimum is always 4√(I₁ I₂).

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