Q: A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on 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.

Sol: $\large \theta = \frac{y}{D} = \frac{2.5 \times 10^{-3}}{1} rad$

Now, d sin θ = n ?

Since θ is very small, therefore sin θ = θ.

or , $\large d = \frac{n \lambda}{\theta} $

$\large d = \frac{1 \times 500 \times 10^{-9}}{2.5 \times 10^{-3}} $

d = 2 × 10^{-4} m = 0.2 mm.