Question

A concave lens of focal length $f$ forms an image whose height is $n$ times the height 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 $f$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{u}{v}-1=\frac{u}{f}$ ---------$1$

Now, height of the image is equal to $n\times$ the height of the object.
$\frac{h_i}{h_o}=n$ ------------$2$
here $h_i$ is the height of the image and $h_o$ is the height of the object.

Magnification of a lens,
$m=\frac{h_i}{h_o}=\frac{v}{u}$
$⟹\frac{v}{u}=n$ ---------($from2$)
$\frac{u}{v}=\frac{1}{n}$

Now, $\frac{u}{v}-1=\frac{u}{f}$ ----($from1$)
$\Rightarrow (\frac{1}{n}-1)=\frac{u}{f}$
$\therefore u=\left(\frac{1-n}{n}\right)f$