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, $\cos \, \theta=\sqrt{\frac{R^{2}-r^{2}}{R}}=\left(1-\frac{r^{2}}{R^{2}}\right)^{^{\frac{1}{2}}}$
$\approx1-\frac{r^{2}}{2R^{2}}$
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
$A=\pi d^{2}=\frac{\pi r^{6}}{4R^{4}}$