A ray of light in the direction $-\left(\hat{i} + \sqrt{3}\hat{j}\right)$ is incident on a plane mirror. After reflection it travels along the direction $\dfrac{1}{2}\left(\hat{i} - \sqrt{3}\hat{j}\right)$. The angle of incidence is
Answer: (C) $60°$
Concept. On reflection the component of the ray along the mirror is unchanged and the normal component reverses, so the difference (reflected $-$ incident) points along the normal.
Step 1. Unit vectors. Incident $\hat a=-\dfrac{\hat i+\sqrt3\hat j}{2}$; reflected $\hat b=\dfrac{\hat i-\sqrt3\hat j}{2}$.
Step 2. Normal. $\hat b-\hat a=\dfrac{\hat i-\sqrt3\hat j}{2}+\dfrac{\hat i+\sqrt3\hat j}{2}=\hat i$, so the normal is along $\hat i$.
Step 3. Angle of incidence. It is the angle between the reversed incident ray $-\hat a$ and the normal: $\cos i=(-\hat a)\cdot\hat i=\dfrac12$, so $i=60^\circ$.
Answer. $60^\circ$ (Option C).
Solution by Sreeraj P, M.Sc Physics
