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 a
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

Answer: (D)

since the rod is infinitely long the image of end $B$ will be formed at focus. Image of $A$ can be found using $\frac{1}{v}+\frac{1}{(-u)}=\frac{1}{-f}$
So $v=\frac{fu}{f-u}$
This $V$ is position of image and we know image of $A$ will be real hence distance of image $=\frac{fu}{u-f}$
so the size of image $=\frac{fu}{u-f}-f$
$=\frac{fu-fu+f^2}{u-f}=\frac{f^2}{u-f}$