In a Young's double-slit experiment, two slits are 1.5 mm apart while the screen is 1.2 m away. When a light of wavelength 600 nm is incident on slits, the fringe width will be

0.48 mm

The correct answer is Option (1) → 0.48 mm

Given:

Distance between slits, $d = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}$

Distance between slits and screen, $D = 1.2 \text{ m}$

Wavelength of light, $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$

Fringe width formula:

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

Substitute values:

$\beta = \frac{600 \times 10^{-9} \times 1.2}{1.5 \times 10^{-3}}$

$\beta = \frac{720 \times 10^{-9}}{1.5 \times 10^{-3}}$

$\beta = 480 \times 10^{-6} \text{ m}$