Question

A concave mirror of radius of curvature $R$ has a circular outline of radius $r.A$ circular disc is to be placed normal to the axis at the focus, so that it collects all the light that is reflected from the mirror from a beam parallel to the axis. For $r<<R,$ the area of this disc has to be a t least

• A. $\frac{\pi r^6}{4R^4}$
• B. $\frac{\pi r^4}{4R^2}$
• C. $\frac{\pi r^5}{4R^3}$
• D. $\frac{\pi r^4}{R^2}$

Answer 1

Answer: (A)

For a small mirror, rays parallel to axis converges at focus. But if apperture of mirror is large, rays reflected from mirror forms a circular (not point) image

Now, consider a ray reflected from periphery of mirror as shown below

In above figure, $C=$ centre of curvature,
$CN=$ normal,
$F=$ focus of mirror,
$P=$ pole of mirror,
$d=$radius of disc
and $r=$ radius of circular section of mirror.
In $\Delta QNC,$ perpendicular $QM$ is drawn.
From $\Delta QMC,$
$QC=\frac{CM}{\cos \theta }=\frac{R}{2\cos \theta }$
So, $PQ=R-\frac{R}{2\cos \theta }$
$\therefore QF=PF-PQ=\frac{R}{2}-\left(R-\frac{R}{2\cos \theta }\right)$
$=\frac{R}{2\cos \theta }-\frac{R}{2}$
Now, from similar triangles $\Delta PQN$ and $\Delta QFL$, we have
$\frac{r}{d}=\frac{pQ}{QF}$
$\Rightarrow \frac{r}{d}=\frac{\left(R-\frac{R}{2\cos \theta }\right)}{\frac{R}{2\cos \theta }-\frac{R}{2}}$
$=\frac{2\cos \theta -1}{1-\cos \theta }\cdots \left(i\right)$
Now, $\cos \theta =\sqrt{\frac{R^2-r^2}{R}}={\left(1-\frac{r^2}{R^2}\right)}^{{}^{\frac{1}{2}}}$
$\approx 1-\frac{r^2}{2R^2}$
Substituting in Eq. $\left(i\right),$ we get
$\frac{r}{d}=\frac{2-\frac{r^2}{R^2}-1}{\frac{r^2}{2R^2}}=\frac{\left(1-\frac{r^2}{R^2}\right)}{\left(\frac{r^2}{2R^2}\right)}$
$\Rightarrow d=\frac{r^3}{2R^2}/\left(1-\frac{r^2}{R^2}\right)$
$\Rightarrow d=\frac{r^3}{2R^2}.\left(1-\frac{r^2}{R^2}\right)$
$\therefore d\approx \frac{r^3}{2R^2}$
So, area of disc required is
$A=\pi d^2=\frac{\pi r^6}{4R^4}$