In a diffraction from a single slit of width 0.30 mm, first order maxima is found with light of wavelength 6000 Å. The angular spread of first order maxima on either side of the central maxima is

$3×10^{-3}\, rad$

The correct answer is Option (2) → $3×10^{-3}\, rad$

For a single slit diffraction, the angular positions of minima are given by:

$a \sin \theta = m \lambda$, $m = 1, 2, 3,...$

The width of the central maximum extends from the first minimum on one side to the first minimum on the other side. For first-order maximum beyond central maximum, approximate angular width is half the angular separation between first minima.

Given:

Slit width $a = 0.30 \, \text{mm} = 3 \times 10^{-4} \, \text{m}$

Wavelength of light: $\lambda = 6000 \, \text{Å} = 6 \times 10^{-7} \, \text{m}$

Angle to first minimum: $\sin \theta \approx \theta = \frac{\lambda}{a} = \frac{6 \times 10^{-7}}{2 \times 10^{-4}} = 3 \times 10^{-3} \, \text{rad}$

Angular spread of first-order maxima on either side of central maximum: $\theta \approx 0.003 \, \text{rad}$