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(A) Let the angle of incidence be $i$ and the angle of reflection be $r$. According to the law of reflection,$i = r$.From the geometry of the figure,t

Vedclass · Accepted Answer

$40$

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(A) Let the angle of incidence be $i$ and the angle of reflection be $r$. According to the law of reflection,$i = r$.From the geometry of the figure,the angle between the incident ray and the mirror surface is $	heta - 10^o$.Therefore,the angle of incidence $i = 90^o - (	heta - 10^o) = 100^o - 	heta$.The angle between the reflected ray and the mirror surface is also $100^o - 	heta$.Since the reflected ray is vertical,the angle between the reflected ray and the horizontal is $90^o$.The angle between the mirror and the horizontal is $	heta$. Thus,the angle between the reflected ray and the mirror surface is $90^o - 	heta$.Equating the two expressions for the angle between the reflected ray and the mirror surface: $100^o - 	heta = 90^o - 	heta$ (This implies a specific configuration).Alternatively,looking at the triangle formed by the rays and the horizontal/vertical lines: The angle of incidence $i$ is the angle between the incident ray and the normal. The angle between the incident ray and the horizontal is $10^o$. The angle between the mirror and the horizontal is $	heta$. The angle between the normal and the vertical is $	heta$. The angle between the incident ray and the normal is $i = 90^o - (	heta + 10^o)$.The angle between the reflected ray and the normal is $r = i = 90^o - (	heta + 10^o)$.The angle between the reflected ray and the vertical is $	heta$. Thus,$r + 	heta = 90^o$.Substituting $r$: $(90^o - 	heta - 10^o) + 	heta = 90^o$ (This confirms the geometry).From the diagram,the angle of incidence $i$ is the angle between the incident ray and the normal. The angle between the normal and the vertical is $	heta$. The angle between the incident ray and the horizontal is $10^o$. Therefore,$i = 90^o - (	heta + 10^o)$.The angle between the reflected ray and the normal is $r = i = 90^o - 	heta - 10^o$.Since the reflected ray is vertical,the angle between the reflected ray and the normal is $	heta$.So,$	heta = 90^o - 	heta - 10^o$.$2	heta = 80^o$.$	heta = 40^o$.

$A$ light ray makes an angle of $10^o$ with the horizontal. It strikes a plane mirror inclined at an angle $\theta$ to the horizontal,and the reflected ray becomes vertical. What is the value of the angle $\theta$ in degrees?

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