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(A) $1$. The incident ray strikes mirror $M_1$ at an angle of incidence $i_1 = 90^{\circ} - 40^{\circ} = 50^{\circ}$.$2$. The reflected ray from $M_1$

Vedclass · Accepted Answer

$5^{\circ}$ clockwise

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(A) $1$. The incident ray strikes mirror $M_1$ at an angle of incidence $i_1 = 90^{\circ} - 40^{\circ} = 50^{\circ}$.$2$. The reflected ray from $M_1$ makes an angle of $40^{\circ}$ with the horizontal.$3$. Let the angle of mirror $M_2$ with the horizontal be $\alpha = 25^{\circ}$. The angle of incidence at $M_2$ is $i_2 = 40^{\circ} + 25^{\circ} = 65^{\circ}$.$4$. For the final reflected ray to be horizontal,the angle of reflection at $M_2$ must be $r_2 = 0^{\circ}$ relative to the horizontal,meaning the ray must be parallel to the horizontal.$5$. If the mirror $M_2$ is rotated by an angle $	heta$,the new angle of the mirror with the horizontal becomes $\alpha'$. The angle of incidence at $M_2$ becomes $i_2' = 40^{\circ} + \alpha'$.$6$. For the reflected ray to be horizontal,the angle of the reflected ray with the mirror must be equal to the angle of incidence. Geometry shows that for the reflected ray to be horizontal,the angle of incidence at $M_2$ must be $20^{\circ}$.$7$. Thus,$40^{\circ} + \alpha' = 65^{\circ} - 	heta = 20^{\circ} \Rightarrow 	heta = 5^{\circ}$ clockwise.

By what angle should $M_2$ be rotated,so that the light ray after reflection from both mirrors becomes horizontal? (The angle of incidence at mirror $M_1$ is $40^{\circ}$ and the angle between $M_2$ and the horizontal is $25^{\circ}$ as shown in the figure.)

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