The mirrors are perpendicular to each other a | vedclass

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(D) Let the incident ray make an angle $i$ with the normal to mirror $M_1$. By the law of reflection, the angle of reflection is also $i$. The angle t

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For any $i$ between $0°$ and $90°$

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(D) Let the incident ray make an angle $i$ with the normal to mirror $M_1$. By the law of reflection, the angle of reflection is also $i$. The angle the reflected ray makes with the mirror $M_1$ is $(90° - i)$. Since the mirrors are perpendicular, the reflected ray makes an angle of $90° - (90° - i) = i$ with the normal to mirror $M_2$. By the law of reflection at $M_2$, the final reflected ray makes an angle $i$ with the normal to $M_2$. The total deviation produced by two mirrors placed at an angle $	heta$ is $\delta = 360° - 2	heta$. Here, $	heta = 90°$, so $\delta = 360° - 2(90°) = 180°$. $A$ deviation of $180°$ means the final ray is anti-parallel to the incident ray. However, if we consider the geometry of the rays, the angle of the final ray with the incident ray is always $180°$ regardless of the angle of incidence $i$, provided both reflections occur. Thus, the final ray is always anti-parallel to the incident ray. Wait, if the question asks for the ray to be parallel, and the deviation is $180°$, it is anti-parallel. In many contexts, 'parallel' is used loosely to mean 'parallel to the original direction'. Given the options, for any angle of incidence $i$ (where both reflections occur), the ray remains anti-parallel. Therefore, the condition holds for any $i$.

Column $I$	Column $II$
$A$. Concave Mirror	$(p)$ Real image
$B$. Convex Mirror	$(q)$ Virtual image
$C$. Convex Lens	$(r)$ Magnified image
$D$. Concave Lens	$(s)$ Image at infinity

The mirrors are perpendicular to each other as shown in the figure. $A$ light ray $AB$ is incident on the mirror $M_1$. The reflected ray then undergoes a reflection from the mirror $M_2$. The final ray after reflection from $M_2$ will be parallel to the incident ray, if

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