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(C) Let the required angle of incidence at the first mirror after reflection from the second mirror be $	heta$.From the geometry of the figure:$1$. A

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(C) Let the required angle of incidence at the first mirror after reflection from the second mirror be $	heta$.From the geometry of the figure:$1$. At the first point of incidence $C$ on the lower mirror,the angle of incidence is $50^{\circ}$. Thus,the angle of reflection is also $50^{\circ}$. The angle the reflected ray makes with the mirror surface is $90^{\circ} - 50^{\circ} = 40^{\circ}$.$2$. In $\Delta ABC$,where $\angle A = 60^{\circ}$ and the angle at $C$ is $40^{\circ}$,the angle $\alpha$ at vertex $B$ is $\alpha = 180^{\circ} - (60^{\circ} + 40^{\circ}) = 80^{\circ}$.$3$. The ray hits the second mirror at $B$. The angle of incidence at $B$ is $\beta = 90^{\circ} - \alpha = 90^{\circ} - 80^{\circ} = 10^{\circ}$.$4$. The ray reflects from $B$ and hits the first mirror again at point $D$. In $\Delta ABD$,the angle at $A$ is $60^{\circ}$. The angle at $B$ is $\angle ABD = \alpha + \beta = 80^{\circ} + 10^{\circ} = 90^{\circ}$ (since the reflected ray makes an angle $\beta$ with the normal,the angle with the mirror is $90^{\circ}-\beta$,and the total angle at $B$ inside the triangle is $\alpha + \beta$).$5$. In $\Delta ABD$,the sum of angles is $180^{\circ}$. Thus,$60^{\circ} + 90^{\circ} + (90^{\circ} - 	heta) = 180^{\circ}$.$6$. Solving for $	heta$: $240^{\circ} - 	heta = 180^{\circ} \implies 	heta = 70^{\circ}$.

$A$ ray of light is incident at $50^{\circ}$ on the middle of one of the two mirrors arranged at an angle of $60^{\circ}$ between them. The ray then touches the second mirror,gets reflected back to the first mirror,making an angle of incidence of.........$^{\circ}$

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