A double slits is illuminated by light of wavelength 600 nm. The slits are 0.1 cm apart and the screen is placed 1 m away. The width of maxima is:

0.6 mm

The correct answer is Option (2) → 0.6 mm

The width of a maxima in a double-slit interference pattern -

$Δy=\frac{λL}{d}$

λ = 600 nm (wavelength of light)

$L=1m$ (Distance between slits & screen)

$d=0.1cm=0.001m$ (distance between slits)

$Δy=\frac{600×10^{-9}×1}{0.001}$

$=0.6mm$