Question

A concave mirror of radius of curvature R has a circular outline of radius r. A circular disk 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 at 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

The correct option is A $\frac{\pi r^6}{4R^4}$

$A=\pi d^2$

from similar triangle

$\frac{r}{d}=\frac{R-\frac{R}{2cos\theta }}{\frac{R}{2}-(R-\frac{R}{2cos\theta })}$

$d=\frac{-cos\theta +1}{2cos\theta -1}r$

also $cos\theta =\frac{\sqrt{R^2-r^2}}{R}$

on solving taking the consideration R>>>r

$A=\pi d^2=\frac{\pi r^6}{4R^4}$