Angular width of a central max. is 30° when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx.
Answer & explanation
Correct answer: option 2
Angular width $=2 \theta=30^{\circ}$
∴ $\sin 30^{\circ}=\frac{\lambda}{a}$
∴ a = $\frac{\lambda}{\sin 30^{\circ}}=\frac{6000 \times 10^{-10}}{1 / 2}=12000 \times 10^{-10}$ m
= 12 × 10–7 m