Question

An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is at distance u > f from the mirror. Its image will have a length:

• A. $\frac{uf}{u-f}$
• B. $\frac{uf}{u+f}$
• C. $\frac{f^2}{u+f}$
• D. $\frac{f^2}{u-f}$

Answer 1

The correct option is D $\frac{f^2}{u-f}$

One end is at infinity,therefore its image will formed at focus F,

For near end, $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{-u}+\frac{1}{v}=\frac{1}{-f}$

$v=\frac{fu}{f-u}=-\left(\frac{fu}{u-f}\right)$ ,$u>f$

Therefore length of image will be,

=$\frac{fu}{u-f}-f$

=$\frac{fu-f(u-f)}{u-f}$

=$\frac{f^2}{u-f}$