Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)

Solution

Physics sir · 3 steps

Intensities add as amplitudes, so take square roots first, then add and subtract them.

1. Intensity ratio. Intensity is proportional to slit width, so I₁ : I₂ = 9 : 16. Take I₁ = 9I and I₂ = 16I.
2. Amplitude ratio. Amplitude goes as the square root of intensity, so a₁ : a₂ = √9 : √16 = 3 : 4.

   Iₘₐₓ/Iₘᵢₙ = (a₁ + a₂)²/(a₁ - a₂)² = (3+4)²/(4-3)² = 49/1

3. So. At a bright fringe the amplitudes add to 7, and at a dark fringe they leave 1. Squaring both gives Iₘₐₓ : Iₘᵢₙ = 49 : 1, option (d).

The idea to keep. In interference, amplitudes add, not intensities. Convert each intensity to √I, add and subtract, then square: Iₘₐₓ/Iₘᵢₙ = ((√I₁+√I₂)/(√I₁-√I₂))².

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