Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

A rod of infinite length is placed 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. The length of it's image is

Options:

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

$\frac{f^2}{u+f}$

$\frac{uf}{u+f}$

$\frac{uf}{u-f}$

Correct Answer:

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

Explanation:

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

$v = \frac{uf}{u - f}$

$\text{For far end: } u \rightarrow \infty \Rightarrow v = f$

$\text{Image extends from } v = \frac{uf}{u - f} \text{ to } v = f$

$\text{Length of image} = \frac{uf}{u - f} - f$

$= f\left(\frac{u}{u - f} - 1\right)$

$= f\left(\frac{u - (u - f)}{u - f}\right)$

$= f\left(\frac{f}{u - f}\right)$

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

The length of the image is $\frac{f^2}{u - f}$.