Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

Solution

Physics sir · 4 steps

Slit width sets intensity, not amplitude. Take square roots first, then add and subtract.

1. Width to intensity. The light passing through a slit is proportional to its width. So if one slit is half as wide, I₁ = 2I and I₂ = I.
2. Intensity to amplitude. Amplitude goes as √I, so the amplitudes are in the ratio a₁ : a₂ = √2 : 1. Waves interfere by adding amplitudes, not intensities.

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

3. Expand the squares. (√2+1)² = 2 + 2√2 + 1 = 3 + 2√2 and (√2-1)² = 3 - 2√2.
4. So. Iₘₐₓ : Iₘᵢₙ = (3+2√2) : (3-2√2), which is option (d).

The idea to keep. In YDSE, slit width tells you intensity. Always convert to amplitudes with a square root, then use ((a₁+a₂)/(a₁-a₂))² for Iₘₐₓ/Iₘᵢₙ.

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