Question

A concave lens of focal length $f$ forms an image which is $n$ times the size of the object. The distance of the object from the lens is:

• A. $(1-n)f$
• B. $(1+n)f$
• C. $\left(\frac{1+n}{n}\right)f$
• D. $\left(\frac{1-n}{n}\right)f$

Answer 1

The correct option is D $\left(\frac{1-n}{n}\right)f$

Let the object distance be $u$, image distance be $v$ and focal length be $f$.
From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{u}{v}-1=\frac{u}{f}$ .......(i)
Let height of the image be $h_i$ and height of the object be $h_o$
Given, $h_i=n\times h_o$
or magnification, $m=\frac{h_i}{h_o}=\frac{v}{u}=n$
$\Rightarrow v=n\times u$
From equation (i) we have,
$\Rightarrow \frac{u}{u\times n}-1=\frac{u}{f}$
$\Rightarrow (\frac{1}{n}-1)=\frac{u}{f}$
$\therefore u=\left(\frac{1-n}{n}\right)f$