A slit is illuminated by light of wavelength 6500 Å. The first minimum is observed at θ = 30°. The width of the slit is

1.3 μm

The correct answer is Option (3) → 1.3 μm

Given:

Wavelength, $\lambda = 6500\ \text{Å} = 6500 \times 10^{-10}\ \text{m} = 6.5 \times 10^{-7}\ \text{m}$

First minimum angle, $\theta = 30^\circ$

For single slit diffraction, condition for first minimum:

$a \sin\theta = \lambda$

Rearranging for slit width:

$a = \frac{\lambda}{\sin\theta} = \frac{6.5 \times 10^{-7}}{\sin 30^\circ} = \frac{6.5 \times 10^{-7}}{0.5} = 1.3 \times 10^{-6}\ \text{m}$

∴ Width of the slit = $1.3 \times 10^{-6}\ \text{m}$