Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is x I. The value of x is___________ .

Solution

Physics sir · 4 steps

Add the amplitudes, not the intensities: take square roots first, then square at the end.

1. Work with amplitudes. Intensity goes as amplitude squared, so amplitude goes as √I. The two amplitudes are in the ratio √4I : √9I = 2 : 3, taking √I as one unit.

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

2. Put in numbers. At a bright fringe the amplitudes add: Iₘₐₓ = (2+3)² I = 25I. At a dark fringe they subtract: Iₘᵢₙ = (3-2)² I = I.
3. Quick check. Expanding both squares, the I₁ + I₂ terms cancel and Iₘₐₓ - Iₘᵢₙ = 4√(I₁ I₂) = 4√36I² = 24I.
4. So. Iₘₐₓ - Iₘᵢₙ = 25I - I = 24I, so x = 24.

The common slip: 12. Taking the difference as 2√(I₁ I₂) instead of 4√(I₁ I₂) gives 12, not 24. The cross term appears with a plus sign in Iₘₐₓ and with a minus sign in Iₘᵢₙ, so it counts twice.

The idea to keep. In interference, amplitudes add and intensities do not. Take √I of each beam, add or subtract, then square. The difference Iₘₐₓ - Iₘᵢₙ is always 4√(I₁ I₂).

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