A ray of light is incident on a concave mirro | vedclass

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Question

(A) Let $C$ be the center of curvature and $F$ be the focus of the concave mirror. Let the incident ray strike the mirror at point $A$. The height of

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) Let $C$ be the center of curvature and $F$ be the focus of the concave mirror. Let the incident ray strike the mirror at point $A$. The height of the incident ray from the principal axis is $AD = f$,where $f$ is the focal length.In $\Delta ADC$,where $D$ is the projection of $A$ on the principal axis,$\sin i = \frac{AD}{AC} = \frac{f}{R} = \frac{f}{2f} = \frac{1}{2}$.Thus,the angle of incidence $i = 30^{\circ}$.Since the reflected ray $AB$ makes an angle $2i$ with the principal axis,in $\Delta ADB$,$\sin 2i = \frac{AD}{AB} = \frac{f}{AB}$.Therefore,$AB = \frac{f}{\sin 60^{\circ}} = \frac{f}{\sqrt{3}/2} = \frac{2f}{\sqrt{3}}$.In $\Delta ABC$,since $CA$ is the normal,$\angle CAB = i = 30^{\circ}$. Also,$\angle ACB = i = 30^{\circ}$ (alternate interior angles). Thus,$\Delta ABC$ is isosceles with $CB = AB = \frac{2f}{\sqrt{3}}$.The distance of the focus from the center of curvature is $FC = R - f = 2f - f = f$.The required ratio is $\frac{CB}{FC} = \frac{2f/\sqrt{3}}{f} = \frac{2}{\sqrt{3}}$.

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