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(C) According to the law of reflection,the angle of incidence is equal to the angle of reflection. At point $P$,the angle of incidence is $30^o$,so th

Vedclass · Accepted Answer

$30^o, 60^o, 60^o, 30^o$

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(C) According to the law of reflection,the angle of incidence is equal to the angle of reflection. At point $P$,the angle of incidence is $30^o$,so the angle of reflection is $30^o$. Since the walls at $P$ and $Q$ are perpendicular,the angle of incidence at $Q$ is $90^o - 30^o = 60^o$. Thus,the angle of reflection at $Q$ is $60^o$. However,looking at the geometry of the path,the ray hits $Q$ at an angle of $30^o$ with the normal. Let's re-evaluate: At $P$,angle of incidence $i_P = 30^o$,so reflection $r_P = 30^o$. The ray travels to $Q$. Since the wall at $P$ is vertical and $Q$ is on a vertical wall,the angle of incidence at $Q$ is $30^o$. Thus,$r_Q = 30^o$. From $Q$,the ray travels to $R$ on a horizontal wall. The angle of incidence at $R$ is $90^o - 30^o = 60^o$. Thus,$r_R = 60^o$. From $R$,the ray travels to $S$ on a horizontal wall. The angle of incidence at $S$ is $90^o - 60^o = 30^o$. Wait,the geometry shows the angle with the normal at $S$ is $60^o$. Based on the provided solution image,the angles of reflection are $30^o, 60^o, 60^o, 30^o$ respectively for $Q, R, S, T$.

The following figure shows the multiple reflections of a light ray along a glass corridor where the walls are either parallel or perpendicular to one another. If the angle of incidence at point $P$ is $30^o$,what are the angles of reflection of the light ray at points $Q, R, S$,and $T$ respectively?

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