A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away.
Question:

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Solution:

Wavelength of light beam, λ = 500 nm = 500 × 10−9 m

Distance of the screen from the slit, D = 1 m

For first minima, n = 1

Distance between the slits = d

Distance of the first minimum from the centre of the screen can be obtained as:

x = 2.5 mm = 2.5 × 10−3 m

It is related to the order of minima as:

$n \lambda=x \frac{d}{D}$

$d=\frac{n \lambda D}{x}$

$=\frac{1 \times 500 \times 10^{-9} \times 1}{2.5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm}$

Therefore, the width of the slits is 0.2 mm.