Question

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

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

Answer 1

Answer: (C)

${70}^o$

Let required angle be $\theta$

From geometry of figure

In $\Delta ABC;\alpha ={180}^o-({60}^o+{40}^o)={80}^o$

$\Rightarrow \beta ={90}^o-{80}^o={10}^o$

In $\Delta ABD,\angle A={60}^o,\angle B=(\alpha +2\beta )$

$=(80+2\times 10)={100}^o$ and $\angle D=({90}^o-\theta )$

$\because \angle A+\angle B+\angle D={180}^o\Rightarrow {60}^o+{100}^o+({90}^o-\theta )={180}^o\Rightarrow \theta$

$={70}^o$