Question

A thin rod of length $\frac{f}{3}$ lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is

• A. f
• B. $\frac{f}{2}$
• C. 2f
• D. $\frac{f}{4}$

Answer 1

The correct option is B $\frac{f}{2}$

If end A of rod acts an object for mirror then it's image will be A' and if $u=2f-\frac{f}{3}=\frac{5f}{3}$
So by using $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\Rightarrow \frac{1}{-f}=\frac{1}{v}+\frac{1}{\frac{-5f}{3}}\Rightarrow v=-\frac{5}{2}f$
$\therefore$ Length of image =$\frac{5}{2}f-2f=\frac{f}{2}$