A parallel beam of light of wavelength 500 nm is incident normally on a slit of width d. If the distance between slit and screen is 0.6 m and distance of second order minimum from the centre of the screen is 6.0 mm, What is the width of the slit?

0.1 mm

The correct answer is Option (1) → 0.1 mm

The minima occur when -

$d\sin θ=mλ$

where,

d = Width of the slit

θ = Angle of diffraction

m = order of minima

λ = wavelength of light

The angle θ for the second order minimum is -

$\tan θ=\frac{y_3}{L}=\frac{6×10^{-3}}{0.6}=0.01$

for small θ,

$\sin θ≃0.01$

$∴d×\sin θ=mλ$

$d×0.01=2×500×10^{-9}$

$⇒d=1cm$