Angle of Incidence for Reflected Light Ray Vector

To find the angle of incidence, let the unit vector in the direction of the incident light ray be: $\hat{u}_i = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$

The unit vector in the direction of the reflected light ray is: $\hat{u}_r = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$

The angle between the incident ray vector $\hat{u}_i$ and the reflected ray vector $\hat{u}_r$, denoted by $\theta$, can be found using the scalar (dot) product: $\cos\theta = \hat{u}_i \cdot \hat{u}_r$

Substituting the given vectors: $\cos\theta = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)$ $\cos\theta = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$

This gives the angle between the directions of the incident and reflected rays: $\theta = 120^\circ$

From the law of reflection, the angle between the direction of the incident ray and the direction of the reflected ray is related to the angle of incidence $i$ by: $\theta = 180^\circ - 2i$

Equating the two expressions: $180^\circ - 2i = 120^\circ$ $2i = 60^\circ \implies i = 30^\circ$

Alternatively, observing that the reflection reverses only the component along the $\hat{j}$ direction while leaving the $\hat{i}$ component unchanged, the normal to the mirror is parallel to the $y$-axis ($\hat{n} = \hat{j}$).

The angle of incidence $i$ is the angle between the incident ray and the normal to the mirror: $\cos i = |\hat{u}_i \cdot \hat{j}| = \left|\frac{\sqrt{3}}{2}\right| = \frac{\sqrt{3}}{2}$ $i = 30^\circ$

Therefore, the correct option is A.