In an interference experiment, the two slits are 1 mm apart and the screen is placed 1 m away from the slits. The width of a bright fringe for the light of wavelength 500 nm is

0.5 mm

The correct answer is Option (3) → 0.5 mm

Fringe width formula: $\beta = \frac{\lambda D}{d}$

Given: $\lambda = 500 \times 10^{-9} \, m$

$D = 1 \, m$

$d = 1 \, mm = 1 \times 10^{-3} \, m$

$\beta = \frac{500 \times 10^{-9} \times 1}{1 \times 10^{-3}}$

$\beta = 5 \times 10^{-4} \, m$

$\beta = 0.5 \, mm$

Final Answer: $0.5 \, mm$