The ratio of intensities at two points P and Q on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are π / 3 and π / 2, respectively are

Solution

Physics sir · 4 steps

Same amplitudes, so intensity goes as cos squared of half the phase difference. Halve the angle first.

1. Intensity formula. When two waves of equal intensity I₀ superpose with phase difference φ, the resultant intensity depends only on φ.

   I = 4I₀cos²(φ/2)

2. Point P. Here φ = π/3, so φ/2 = π/6. Then cos²(π/6) = 3/4, which gives I_P = 4I₀ × 3/4 = 3I₀.
3. Point Q. Here φ = π/2, so φ/2 = π/4. Then cos²(π/4) = 1/2, which gives I_Q = 4I₀ × 1/2 = 2I₀.
4. So. I_P : I_Q = 3I₀ : 2I₀ = 3 : 2, which is option (d).

The idea to keep. For two equal-amplitude waves, I = 4I₀cos²(φ/2). Always take half the phase difference before squaring the cosine; 4I₀ cancels in any ratio.

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