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A ray of light is incident on a concave mirror

Question : A ray of light is incident on a concave mirror . It is parallel to the principal axis and its height from principal axis is equal to the focal length of the mirror. The ratio of the distance of point B to the distance of the focus from the centre of curvature is : (AB is the reflected ray)

(1) $\displaystyle \frac{2}{\sqrt{3}}$

(2) $\displaystyle \frac{\sqrt{3}}{2}$

(3) $\displaystyle \frac{2}{3}$

(4) $\displaystyle \frac{1}{2}$

Solution :

In ΔADC,

$\displaystyle sin\theta =\frac{CD}{CA}=\frac{f}{R}=\frac{f}{2f}=\frac{1}{2}$

$\displaystyle \Rightarrow \theta =30{}^\circ$

Now the ΔABC is an isosceles triangle, so

CB = BA

In ΔABE,

$\displaystyle sin2\theta =\frac{BE}{BA}=sin60{}^\circ$

$\displaystyle \Rightarrow BA=\frac{BE}{sin60{}^\circ }=\frac{f}{{}^{\sqrt{3}}\!\!\diagup\!\!{}_{2}\;}=\frac{2f}{\sqrt{3}}$

$\displaystyle \Rightarrow BA=CB=\frac{2f}{\sqrt{3}}$

Required ratio,

$\displaystyle \frac{CB}{CF}=\frac{\frac{2f}{\sqrt{3}}}{f}=\frac{2}{\sqrt{3}}$

Hence correct option is (1).

Note :-

In this example, you might initially think that any ray parallel to the principal axis must pass through the focus, so point B should coincide with the focus F. If that were true, we would get

$\displaystyle \frac{CB}{CF}=\frac{f}{f}=1$

However, this is true only for paraxial rays (rays very close to the principal axis). In this question, the incident ray is a marginal ray (its height from the axis is comparable to the focal length), so the paraxial approximation does not hold. As a result, point B does not coincide with the focus F — which is why we must solve geometrically to find the actual point of intersection.

A ray of light is incident on a concave mirror

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