question_answer

A ray of light is incident at 50° on the middle of one of the two mirrors arranged at an angle of 60° between them. The ray then touches the second mirror, get reflected back to the first mirror, making an angle of incidence of [MP PET 2005]

A) \[50{}^\circ \]

B) \[60{}^\circ \]

C) \[70{}^\circ \]

D) \[80{}^\circ \]

Correct Answer: C

Solution :

Let required angle be q     From geometry of figure \[In\,\Delta \,A\,B\,C\,;\,\,a=180{}^\circ \left( 60{}^\circ +40{}^\circ  \right)=80{}^\circ \] \[\Rightarrow \,\,\beta ~=90{}^\circ 80{}^\circ =10{}^\circ \] \[In\,\Delta \,A\,B\,D\,;\,\angle A=60{}^\circ ,\angle B=(a+2\beta )\] \[=(80+2\times 10)=100{}^\circ \,and\,\angle D=(90{}^\circ \theta )\] \[\because \,\angle A+\angle B+\angle D=180{}^\circ \] \[\Rightarrow \,\,60{}^\circ +100{}^\circ +(90{}^\circ \theta )=180{}^\circ \,\,\Rightarrow \theta =70{}^\circ \]