In Young's double slit experiment, slits are illuminated by light of wavelength 650 nm. The slits are 0.13 cm apart and the screen is placed 1.2 m away. What will be the separation of the adjacent maxima?

0.6 mm

The correct answer is Option (4) → 0.6 mm

Given:

$\lambda = 650 \, nm = 650 \times 10^{-9} \, m$

$d = 0.13 \, cm = 0.13 \times 10^{-2} = 1.3 \times 10^{-3} \, m$

$D = 1.2 \, m$

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

$\beta = \frac{650 \times 10^{-9} \times 1.2}{1.3 \times 10^{-3}}$

$\beta = \frac{7.8 \times 10^{-7}}{1.3 \times 10^{-3}}$

$\beta = 6.0 \times 10^{-4} \, m$

$\beta = 0.6 \, mm$

Answer: 0.6 mm