In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Solution

Physics sir · 3 steps

Amplitudes add at a bright fringe and subtract at a dark one, so square the sum and the difference.

1. Amplitude ratio. The question says amplitude is proportional to slit width. One slit is three times wider, so the amplitudes are a₁ = 3a and a₂ = a.
2. Bright and dark fringes. At a maximum the two waves arrive in phase, so the resultant amplitude is a₁ + a₂ = 4a. At a minimum they arrive out of phase, so it is a₁ - a₂ = 2a. Intensity goes as the square of the resultant amplitude.

   Iₘₐₓ/Iₘᵢₙ = (a₁ + a₂)²/(a₁ - a₂)² = (4a)²/(2a)² = 16/4 = 4

3. So. The ratio of maximum to minimum intensity is 4 : 1, which is option (a).

The idea to keep. In interference, combine amplitudes first, then square. Iₘₐₓ/Iₘᵢₙ = (a₁+a₂)²/(a₁-a₂)². Always check whether the question gives you amplitudes or intensities.

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