Question

A concave lens of focal length $f$ forms an image which is $\frac{1}{3}$ times the size of the object. Then, the distance of object from the lens is:

• A. $2f$
• B. $f$
• C. $\frac{2}{3}f$
• D. $\frac{3}{2}f$

Answer 1

The correct option is A $2f$

Magnification due to thin lens $m=\frac{h_i}{h_o}=-\frac{v}{u}$

where all the symbols have their usual meaning.

Given : $\frac{h_i}{h_o}=\frac{1}{3}$

As the magnification by a concave lens is less than 1, thus the image is formed in front of lens. Also the focal length of the concave lens is negative.

$\therefore$ $\frac{1}{3}=-\frac{(-v)}{(-x)}$ $⟹v=\frac{-x}{3}$ where $x$ is the distance of object from lens

Using, $\frac{1}{v}-\frac{1}{u}=\frac{1}{F}$

$\therefore$ $\frac{-3}{x}-\frac{1}{-x}=\frac{1}{-f}$

or, $\frac{-3}{x}+\frac{1}{x}=\frac{-1}{f}$

or, $\frac{-2}{x}=\frac{-1}{f}$ $⟹x=2f$