Step-by-step derivation:

1. Identify the direction vectors of the incident and reflected rays. The problem provides the unit direction vector for the incident ray as: $\hat{I} = \frac{1}{2}(\hat{i} + \sqrt{3}\hat{j})$ And the unit direction vector for the reflected ray as: $\hat{R} = \frac{1}{2}(\hat{i} - \sqrt{3}\hat{j})$

2. Determine the normal to the plane mirror. According to the vector form of the law of reflection, the reflected ray vector $\hat{R}$ is related to the incident ray vector $\hat{I}$ and the unit normal vector $\hat{n}$ by the formula: $\hat{R} = \hat{I} - 2(\hat{I} \cdot \hat{n})\hat{n}$ Rearranging this equation gives: $\hat{I} - \hat{R} = 2(\hat{I} \cdot \hat{n})\hat{n}$ This shows that the vector $\hat{I} - \hat{R}$ is parallel to the normal vector $\hat{n}$. We can calculate this vector difference: $\hat{I} - \hat{R} = \left( \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \right) - \left( \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \right)$ $\hat{I} - \hat{R} = \left( \frac{1}{2} - \frac{1}{2} \right)\hat{i} + \left( \frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) \right)\hat{j}$ $\hat{I} - \hat{R} = 0\hat{i} + \sqrt{3}\hat{j} = \sqrt{3}\hat{j}$ Since $\hat{I} - \hat{R}$ is in the direction of $\hat{j}$, the unit normal vector $\hat{n}$ must be $\hat{j}$ (or $-\hat{j}$). Let's choose $\hat{n} = \hat{j}$.

3. Calculate the angle of incidence. The angle of incidence, $\theta_i$, is the angle between the incident ray vector $\hat{I}$ and the normal vector $\hat{n}$. It is defined as the acute angle between these two directions. We use the dot product formula to find the angle $\theta$ between $\hat{I}$ and $\hat{n}$: $\cos \theta = \frac{\hat{I} \cdot \hat{n}}{|\hat{I}| |\hat{n}|}$ Since $\hat{I}$ and $\hat{n}$ are unit vectors, $|\hat{I}| = 1$ and $|\hat{n}| = 1$. So, $\cos \theta = \hat{I} \cdot \hat{n}$. $\cos \theta = \left( \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \right) \cdot (\hat{j})$ $\cos \theta = \frac{\sqrt{3}}{2}$ Solving for $\theta$, we get: $\theta = \arccos\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$ Since this angle is acute ($0^\circ \le \theta \le 90^\circ$), it is the angle of incidence. $\theta_i = 30^\circ$

4. Verification (Optional) We can verify this by calculating the angle of reflection $\theta_r$ which is the angle between the reflected ray $\hat{R}$ and the normal $\hat{n}$. $\cos \theta' = \hat{R} \cdot \hat{n} = \left( \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \right) \cdot (\hat{j}) = -\frac{\sqrt{3}}{2}$ This gives $\theta' = 150^\circ$. The acute angle is $180^\circ - 150^\circ = 30^\circ$. So, $\theta_r = 30^\circ$. Since $\theta_i = \theta_r$, the result is consistent with the law of reflection.

Final Answer: The angle of incidence is $30^\circ$.