Light of wavelength 600 nm is incident normally on a slit of width 0.03 cm. The width of the central maximum on the screen placed 1 m away from the slit is

4 mm

The correct answer is Option (3) → 4 mm

For a single slit, the angular width of the central maximum is given by:

$\theta = \frac{\lambda}{a}$

Given: $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$, $a = 0.03 \, \text{cm} = 3 \times 10^{-4} \, \text{m}$, $L = 1 \, \text{m}$

Width of central maximum on screen:

$w = 2 L \tan \theta \approx 2 L \theta$ (for small angles)

$\theta = \frac{\lambda}{a} = \frac{600 \times 10^{-9}}{3 \times 10^{-4}} = 2 \times 10^{-3} \, \text{rad}$

$w = 2 \cdot 1 \cdot 2 \times 10^{-3} = 4 \times 10^{-3} \, \text{m}$

w $\approx 4 \, \text{mm}$