A ray of light is incident on a plane reflecting surface at an angle of . Find the deviation in the

A ray of light is incident on a plane reflecting surface at an angle of $30^{\circ}$ . Find the deviation in the incident ray. What will be the deviation if the ray suffers a reflection again at a surface inclined at $60^{\circ}$ to the first surface?
Option: 1
$120^{\circ}$

Option: 2
$180^{\circ}$

Option: 3
$210^{\circ}$

Option: 4
$240^{\circ}$

Answers (1)

$\angle \mathrm{B}^{\prime} \mathrm{CM}_{2}=60^{\circ}, \angle \mathrm{M}_{2} \mathrm{CD}=60^{\circ}, \angle \mathrm{B}^{\prime} \mathrm{CD}=120^{\circ}=\mathrm{D}_{2}$ ,
$\angle \mathrm{DCN}_{2}=\angle \mathrm{N}_{2} \mathrm{CB}=30^{\circ}$

$\angle \mathrm{ABN}_1=\angle \mathrm{N}_1 \mathrm{BC}=30^{\circ} ; \quad \angle \mathrm{CBA}^{\prime}=120^{\circ}=\mathrm{D}_1$

The incident ray of light $\mathrm{A B}$ on the first mirror $\mathrm{M_1}$ at an angle of incidence of $\mathrm{30^{\circ}}$ is reflected along $\mathrm{\mathrm{BC}}$ at an angle of reflection $\mathrm{30^{\circ}}$ .

$\therefore$ Deviation in incedent ray
$=60^{\circ}+60^{\circ}=120^{\circ}$
The reflected ray BC is incident on the second mirror $\mathrm{M_{2}}$ .
Since $\mathrm{\angle \mathrm{M}_1 \mathrm{OM}_2=60^{\circ}}$ ,
$\mathrm{\angle \mathrm{BCO}=60^{\circ}}$ ,

$\mathrm{\therefore \quad \angle \mathrm{BCN}_2=\angle \mathrm{N}_2 \mathrm{CD}=30^{\circ}}$
$\mathrm{\therefore }$ Deviation $\mathrm{D_{2}}$ of the ray
$\mathrm{=60^{\circ}+60^{\circ}=120^{\circ}}$
Total deviation due to the two mirrors
$\mathrm{=D_1+D_2=240^{\circ}}$