A ray of light making an angle 10^o with the | vedclass

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(C) Let the incident ray make an angle of $10^o$ with the horizontal. The reflected ray is vertical, meaning it makes an angle of $90^o$ with the hori

Vedclass · Accepted Answer

$40$

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(C) Let the incident ray make an angle of $10^o$ with the horizontal. The reflected ray is vertical, meaning it makes an angle of $90^o$ with the horizontal.The angle between the incident ray and the reflected ray is the total deviation $\delta$.Since the incident ray is $10^o$ below the horizontal and the reflected ray is $90^o$ above the horizontal, the angle between them is $\delta = 90^o + 10^o = 100^o$.The law of reflection states that the angle of deviation $\delta$ is related to the angle of incidence $i$ by $\delta = 180^o - 2i$.Therefore, $100^o = 180^o - 2i$, which gives $2i = 80^o$, or $i = 40^o$.The angle of incidence $i$ is the angle between the incident ray and the normal to the mirror.Let $	heta$ be the angle the mirror makes with the horizontal. The angle between the incident ray and the mirror surface is $i = 90^o - (	ext{angle between incident ray and mirror surface})$.From the geometry, the angle between the incident ray and the mirror surface is $	heta + 10^o$.Thus, the angle of incidence $i = 90^o - (	heta + 10^o)$.Substituting $i = 40^o$, we get $40^o = 90^o - 	heta - 10^o$.$40^o = 80^o - 	heta$, which implies $	heta = 40^o$.

$A$ ray of light making an angle $10^o$ with the horizontal is incident on a plane mirror making an angle $\theta$ with the horizontal. What should be the value of $\theta$ so that the reflected ray goes vertically upwards?

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