Question

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, get reflected back to the first mirror, making an angle of incidence of

• A. $50^{\circ }$
• B. $60^{\circ }$
• C. $70^{\circ }$
• D. $80^{\circ }$

Answer 1

Answer: (C)

Let required angle be $\theta$

From geometry of figure In
$\Delta ABC;\alpha =180^{\circ }-\left(60^{\circ }+40^{\circ }\right)=80^{\circ }$
$\Rightarrow \beta =90^{\circ }-80^{\circ }=10^{\circ }$
In $\Delta ABD;\angle A=60^{\circ },\angle B=(\alpha +2\beta )$
$=(80+2\times 10)=100^{\circ }$ and $\angle D=\left(90^{\circ }-\theta \right)$
$\because \angle A+\angle B+\angle D=180^{\circ }$
$\Rightarrow 60^{\circ }+100^{\circ }+\left(90^{\circ }-\theta \right)=180^{\circ }$
$\Rightarrow \theta =70^{\circ }$