Question

A $2.0cm$ tall object is placed perpendicular to the principal axis of a concave lens of focal length $15cm$. At what distance from the lens, should the object be placed so that it forms an image $10cm$ from the lens? Also, find the nature and the size of the image formed.

Answer 1

According to the question:
Image distance, $v=-10cm$
Focal length, $f=-15cm$ (since, concave lens)

Let the object distance be $u$.

By lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{-10cm}-\frac{1}{u}=\frac{1}{-15cm}$
$\Rightarrow \frac{1}{u}=-\frac{1}{-15cm}-\frac{1}{10cm}$
$\Rightarrow \frac{1}{u}=\frac{2-3}{30cm}=\frac{-1}{30cm}$
$\Rightarrow \frac{1}{u}=\frac{-1}{30cm}$
$\therefore u=-30cm.$
Thus, object should be placed at $30cm$ in front of lens, in the same side as image.

Now,
Height of object, $h_1=2cm$
Magnification, $m=\frac{h_2}{h_1}=\frac{v}{u}$

Putting value of $v$ and $u$:
Magnification, $m=\frac{h_2}{2cm}=\frac{-10cm}{-30cm}$
$\Rightarrow \frac{h_2}{2cm}=\frac{1}{3}$
$\Rightarrow h_2=\frac{2}{3}cm=0.67cm$
Height of image is $0.67cm$

Thus, the image is virtual, diminished, and erect and one-third the size of object.