In a Young's double slit experiment, the distance between slits is 2.5 mm and the screen is placed 2 m away from the slits. The fringe width for a light of wavelength 400 nm is

0.32 mm

The correct answer is Option (4) → 0.32 mm

Given:

Distance between slits: $d = 2.5~\text{mm} = 2.5 \times 10^{-3}~\text{m}$

Distance to screen: $L = 2~\text{m}$

Wavelength of light: $\lambda = 400~\text{nm} = 400 \times 10^{-9}~\text{m}$

Fringe width in Young's double slit experiment:

$\beta = \frac{\lambda L}{d}$

Substitute values:

$\beta = \frac{400 \times 10^{-9} \cdot 2}{2.5 \times 10^{-3}} = \frac{8 \times 10^{-7}}{2.5 \times 10^{-3}}$

$\beta = 3.2 \times 10^{-4}~\text{m} = 0.32~\text{mm}$

Answer: $\beta = 0.32~\text{mm}$