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