The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

Solution

Physics sir · 3 steps

Slit width sets intensity, not amplitude: take the square root first, then add and subtract amplitudes.

1. Width to intensity. A wider slit lets through more light, so the intensity from a slit is proportional to its width. With widths in the ratio 4:1, the intensities are I₁ : I₂ = 4 : 1.
2. Intensity to amplitude. Intensity goes as amplitude squared, so a₁ : a₂ = √4 : √1 = 2 : 1. Take a₁ = 2 and a₂ = 1.

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

3. So. At a bright fringe the amplitudes add to 3, and at a dark fringe they subtract to 1. Squaring both gives Iₘₐₓ : Iₘᵢₙ = 9 : 1, which is option (d).

The idea to keep. In interference, amplitudes add or subtract, never intensities. Convert width or intensity ratios to amplitudes with a square root, then use (a₁ ± a₂)².

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