Q 12-10-125JEE MainJEE Main 2020 (7 Jan, Shift 1)Easy
Visible light of wavelength $6000\times10^{-8}$ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^\circ$ from the central maximum. If the first minimum is produced at $\theta_1$, then $\theta_1$ is close to:
Answer: (C) $25^\circ$
Minima: $a\sin\theta = n\lambda$.
Second minimum: $a\sin60^\circ = 2\lambda$. First minimum: $a\sin\theta_1 = \lambda$.
$$\sin\theta_1 = \frac{\sin60^\circ}{2} \approx 0.433 \Rightarrow \theta_1 \approx 25.7^\circ \approx 25^\circ$$
Solution by Sreeraj P, M.Sc Physics