Question

A slit of width 'a' is illuminated by a monochromatic light of wavelength 700 nm at normal incidence. Calculate the value of 'a' for position of (i) first minimum at an angle of diffraction of 30. (ii) first maximum at an angle of diffraction of 30

Answer 1

Given

Wavelength, λ=700 nm

Angle of diffraction, θ=300

Width of slit = a

We know that

(i) For destructive interference (minima) in single slit

a sinθ=nλ, for 1st minima n=1⇒ a sinθ=λ⇒a=λsinθ=700 nmsin 300=700 nm1/2=1400 nm

(ii) For constructive interference (maxima) in single slit,

a sinθ=(n+12)λ

In single slit, first maxima occurs at the perpendicular bisector of the slit (n = 0 for central maxima),since in this case occurring at diffraction angle θ so it must be of other order , let n = 1 (say)

⇒a sinθ=3λ2⇒a=3λ2 sinθ=3×700 nm2×sin300=2100 nm

hope it helps pls mark brainliest