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 is the distance of point B to the distance of the focus from the centre of curvature is :

(AB is the reflected ray).

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{3}$
• D. $1$

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}$

Given: 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. AB is the reflected ray.

To find the ratio of the distance of point B from center of curvature to the distance of the focus from the centre of curvature.

Solution:

From above fig,

Let C be the center of curvature and $AB=x$

$AD=f$ and $AC=R$

In $\Delta ADC$,

$\sin i=\frac{AD}{AC}⟹\sin i=\frac{f}{R}⟹\sin i=\frac{f}{2f}⟹\sin i=\frac{1}{2}⟹i={30}^{\circ }........\left(i\right)$

In $\Delta ADB$,

$\sin 2i=\frac{AD}{AB}⟹\sin 2i=\frac{f}{x}.............\left(ii\right)$

As CD is parallel to incident ray, hence made by incident ray with AB is equal to $\angle ABD$, refer above fig.

CA is angle bisector, so $\angle CAB=i$

In $\Delta ABC$,

$\angle ACB=\angle CAB=i⟹AB=CB=x.......\left(iii\right)$

Hence, $CB=x=\frac{f}{\sin 2i}.......\left(iv\right)$ from eqn (ii) and (iii)

The distance of the focus from the centre of curvature, $FC=f.....\left(v\right)$

Now the ratio of the distance of point B from center of curvature to the distance of the focus from the centre of curvature is,

$=\frac{CB}{FC}$

substituting the respective values from eqn (iv) and (v), we get

$⟹\frac{CB}{FC}=\frac{x}{f}⟹=\frac{\frac{f}{\sin 2i}}{f}⟹=\frac{1}{\sin 2i}$

$=\frac{1}{\sin (2\times 30)}$ (from eqn (i))

$=\frac{1}{\frac{\sqrt{3}}{2}}$

Hence the required ratio is $\frac{2}{\sqrt{3}}$