A single slit of width a is illuminated by a monochromatic light of wavelength 600 nm. The value of ' a ' for which first minimum appears at θ=30° on the screen will be :

Solution

Physics sir · 3 steps

Single-slit first minimum: a sin θ = λ, and sin 30° = 0.5 makes a = 2λ.

1. Condition for minima. Split the slit into two halves. At the first minimum, each wavelet from the top half cancels its partner from the bottom half. This happens when the path difference between the two edges of the slit is one full wavelength.

   a sin θ = nλ, n = 1 for the first minimum

2. Put in numbers. With sin 30° = 0.5, a = λ/(sin θ) = (600 nm)/0.5 = 1200 nm.
3. So. a = 1200 nm = 1.2 μm, which is option (d).

The idea to keep. For a single slit, dark fringes sit at a sin θ = nλ and secondary maxima roughly at a sin θ = (n + 1/2)λ. Do not swap in the double-slit conditions.

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