Visible light of wavelength 6000 × 10⁻⁸ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60ᵒ from the central maximum. If the first minimum is produced at θ₁, then θ₁, is close to :

Solution

Physics sir · 3 steps

Minima go by sine, not by angle: halve sin 60°, then find the angle.

1. Condition for minima. For a single slit of width a, the nth dark fringe sits where a sin θₙ = nλ. So sin θₙ grows in equal steps of λ/a. The angle itself does not.

   (sin θ₁)/(sin θ₂) = 1/2

2. Put in numbers. sin θ₁ = 1/2 sin 60° = √3/4 ≈ 0.433. The wavelength cancels, so its value is not needed.
3. So. Since sin 25° ≈ 0.423 and sin 26° ≈ 0.438, θ₁ ≈ 25.7°. That is close to 25°, option (c).

The idea to keep. In single-slit diffraction, sin θ is proportional to the order of the minimum. Take ratios of sines, and use angle ratios only when the angles are small.

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