In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 A is incident on the slit. The first secondary maximum is observed at an angle of

Question

Accepted Answer

(3) sin⁻¹(3/4). Given: first minimum at 30° for λ=5000 A. Idea: minima sit at a sin θ=nλ and secondary maxima roughly midway between them, at a sin θ=(n+1/2)λ. First minimum, n=1: a sin 30°=λ, so a×1/2=λ and a=2λ. First secondary maximum: a sin θ₁=(3λ)/2. sin θ₁=(3λ)/(2a)=(3λ)/(4λ)=3/4. θ₁=sin⁻¹(3/4). The wavelength cancels, so its value is not needed.

In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 A is incident on the slit. The first secondary maximum is observed at an angle of

Step-by-step solution

Why the other options are wrong