Question

In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at θ = 30°. Calculate the width of the slit.

Answer 1

Given: λ = 600 nm

θ = 30°

Formula: d sin θ = nλ, where d is the slit width.

For the first minima in diffraction due to a single slit:

d sin θ = λ    ...(n = 1)

d = `lambda/(sin theta)`

= `(600 xx 10^-9)/(sin 30^\circ)`

= `(600 xx 10^-9)/0.5`

= 1.2 × 10⁻⁶ m

= 1.2 μm