The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1ˢᵗ secondary maxima will be :

Solution

Physics sir · 3 steps

A secondary maximum spans the gap between two neighbouring minima, so its width is λ f / a.

1. Where minima fall. For a single slit, the dark fringes are at a sin θ = nλ. On the focal plane of the lens, the nth minimum is at yₙ = (nλ f)/a from the centre.
2. Width of a maximum. The 1st secondary maximum lies between the 1st and 2nd minima, so its width is y₂ - y₁. This spacing is the same for every secondary maximum, and it is half the width of the central maximum.

   β = (λ f)/a = (400 × 10⁻⁹ m)(1 m)/(0.2 × 10⁻³ m)

3. So. β = 2 × 10⁻³ m = 2 mm, which is option (a).

The idea to keep. In single-slit diffraction, every secondary maximum is λ f / a wide. Only the central maximum is twice as wide, 2λ f / a. Convert everything to metres before dividing.

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