In a Young's double slit experiment, the two parallel slits are 1 mm apart and a screen is placed 1 m away from the slits. When the light of wavelength 600 nm is used, the fringe width is

0.6 mm

The correct answer is Option (3) → 0.6 mm

Fringe width in Young's double slit experiment: $\beta = \frac{\lambda L}{d}$

Given: $\lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, m$, $L = 1 \, m$, $d = 1 \, \text{mm} = 1 \times 10^{-3} \, m$

$\beta = \frac{6 \times 10^{-7} \cdot 1}{1 \times 10^{-3}} = 6 \times 10^{-4} \, m = 0.6 \, mm$

Fringe width β = 0.6 mm