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Question

(C) The incident ray travels in the negative $y$-direction,so its direction vector is $-\hat{j}$.The mirror makes an angle of $30^{\circ}$ with the $Y

Vedclass · Accepted Answer

$\sqrt{3}\hat{i}-\hat{j}$

Vedclass · Answer

(C) The incident ray travels in the negative $y$-direction,so its direction vector is $-\hat{j}$.The mirror makes an angle of $30^{\circ}$ with the $Y$-axis. The normal to the mirror makes an angle of $30^{\circ}$ with the $X$-axis (or $60^{\circ}$ with the $Y$-axis).According to the law of reflection,the angle of incidence equals the angle of reflection.The incident ray makes an angle of $30^{\circ}$ with the mirror surface. Thus,the reflected ray also makes an angle of $30^{\circ}$ with the mirror surface.Looking at the geometry,the reflected ray makes an angle of $30^{\circ}$ with the $X$-axis in the fourth quadrant.The direction vector of the reflected ray can be written as $\vec{v} = v_x \hat{i} + v_y \hat{j}$.Since it is in the fourth quadrant,$v_x > 0$ and $v_y The angle with the positive $X$-axis is $-30^{\circ}$.Thus,the direction is proportional to $(\cos(-30^{\circ})\hat{i} + \sin(-30^{\circ})\hat{j}) = (\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j})$.Multiplying by $2$,we get the vector $\sqrt{3}\hat{i} - \hat{j}$.Therefore,the correct option is $(c)$.

$A$ mirror is placed at an angle of $30^{\circ}$ with respect to the $Y$-axis (see figure). $A$ light ray travelling in the negative $y$-direction strikes the mirror. The direction of the reflected ray is given by the vector

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