Question

A concave lens has a focal length of $20cm$. At what distance from a lens a $5cm$ tall object should be placed, so that it forms an image at $15cm$ from the lens? Also, calculate the size of the image formed.

Answer 1

According to the question,
Image distance, $v=-15cm$ (negative due to convex lens)
Focal length, $f=-20cm$

Let the object distance be $u$.

By lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{-15cm}-\frac{1}{u}=\frac{1}{-20cm}$
$\Rightarrow \frac{1}{u}=\frac{1}{20cm}+\frac{1}{-15cm}$
$\Rightarrow \frac{1}{u}=\frac{3-4}{60cm}=\frac{-1}{60cm}$
$\Rightarrow u=-60cm$
Therefore, object is placed at $60cm$ away from the lens, on the same side as image.

Now,
Height of object, $h_1=5cm$
Magnification, $m=\frac{h_2}{h_1}=\frac{v}{u}$

Putting value of $v$ and $u$:
Magnification, $m=\frac{h_2}{5cm}=\frac{-15cm}{-60cm}$
$\Rightarrow \frac{h_2}{5cm}=\frac{1}{4}$
$\Rightarrow h_2=\frac{5cm}{4}=1.25cm$

Thus, the height of the image is $1.25cm$ and the positive sign means the image is virtual and erect.