Q 12-09-018JEE MainIIT JEE 2013Medium
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) $60°$
Unit vectors: incident $\hat{e}_i = -\dfrac{1}{2}\left(\hat{i} + \sqrt{3}\hat{j}\right)$, reflected $\hat{e}_r = \dfrac{1}{2}\left(\hat{i} - \sqrt{3}\hat{j}\right)$.
The normal to the mirror is along $\hat{e}_r - \hat{e}_i = \dfrac{1}{2}\hat{i} + \dfrac{1}{2}\hat{i} = \hat{i}$.
The angle of incidence $\theta$ is between the reversed incident ray and the normal:
$$\cos\theta = \left(-\hat{e}_i\right)\cdot\hat{i} = \frac{1}{2} \;\Rightarrow\; \theta = 60°$$
Solution by Sreeraj P, M.Sc Physics