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(C) Let the incident unit vector be $\vec{a} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$ and the reflected unit vector be $\vec{b} = \frac{1}{2}

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the incident unit vector be $\vec{a} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$ and the reflected unit vector be $\vec{b} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$.The angle $	heta$ between these two vectors is given by the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$.Since these are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.$\vec{a} \cdot \vec{b} = (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2}) = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$.Therefore,$\cos 	heta = -\frac{1}{2}$,which implies $	heta = 120^{\circ}$.The angle between the incident ray and the reflected ray is $120^{\circ}$.The angle of deviation $\delta = 180^{\circ} - 120^{\circ} = 60^{\circ}$.The angle of deviation is also given by $\delta = 180^{\circ} - 2i$,where $i$ is the angle of incidence.$60^{\circ} = 180^{\circ} - 2i \implies 2i = 120^{\circ} \implies i = 60^{\circ}$.Alternatively,looking at the vectors,the angle with the $y$-axis (normal) is $\cos^{-1}(\frac{\sqrt{3}}{2}) = 30^{\circ}$ if the mirror is along the $x$-axis,but the vectors indicate the normal is the $y$-axis. The angle with the normal is $60^{\circ}$.

$A$ ray of light travelling in the direction $\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection,it travels along the direction $\frac{1}{2}(\hat{i}-\sqrt{3} \hat{j})$. The angle of incidence is: (in $^{\circ}$)

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