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(C) Let the angle of incidence on mirror $M_1$ be $i_1 = 30^o$. The angle of reflection is also $30^o$. The angle the reflected ray makes with the sur

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$4$

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(C) Let the angle of incidence on mirror $M_1$ be $i_1 = 30^o$. The angle of reflection is also $30^o$. The angle the reflected ray makes with the surface of $M_1$ is $90^o - 30^o = 60^o$.In the triangle formed by the two mirrors and the ray,the angle at the vertex is $40^o$. The angle at the base on $M_1$ is $60^o$. Thus,the angle at the base on $M_2$ is $180^o - (40^o + 60^o) = 80^o$.The angle of incidence on $M_2$ is $90^o - 80^o = 10^o$. Since $10^o The angle of reflection from $M_2$ is $10^o$. The angle this ray makes with the surface of $M_2$ is $90^o - 10^o = 80^o$.Now,consider the triangle formed by the ray and the mirrors. The angle at the vertex is $40^o$. The angle at the base on $M_2$ is $80^o$. The angle at the base on $M_1$ is $180^o - (40^o + 80^o) = 60^o$.Since the angle with the surface of $M_1$ is $60^o$,the angle of incidence is $90^o - 60^o = 30^o$. This is the same as the initial angle of incidence,meaning the ray will continue to reflect between the mirrors.However,the question asks for the number of reflections. Based on the geometry,the ray will strike $M_1$ again,then $M_2$,and so on. Given the mirrors are long,the ray will undergo a finite number of reflections until it exits the system. For an angle of $40^o$,the total number of reflections is calculated as $N = \frac{180^o}{	heta} = \frac{180^o}{40^o} = 4.5$. The ray will undergo $4$ reflections.

Two long plane mirrors $M_1$ and $M_2$ are kept inclined to each other. The angle between reflecting surfaces is $40^o$. $A$ light ray strikes mirror $M_1$ at an angle of $30^o$ with the normal. The maximum number of reflections the ray will undergo is:

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