In Young's double slit experiment, two slits are made 4 mm apart and the screen is placed 3 m away from slits. The fringe width when light of wavelength 600 nm is used will be

0.45 mm

The correct answer is Option (2) → 0.45 mm

The fringe width in Young's double slit experiment -

$β=\frac{λD}{d}=\frac{(600×10^{-9})×3}{4×10^{-3}}$

$β=0.45mm$