Question

A convex lens has a focal length of $10cm$. At what distance from the lens should the object be placed so that it forms a real and inverted image $20cm$ away from the lens? What would be the size of the image formed if the object is $2cm$ high? With the help of a ray diagram, show the formation of the image by the lens in this case.

Answer 1

According to the question:

Image distance, $v=20cm$
Focal length, $f=10cm$ (positive since convex lens)

Let the object distance be $u$

By lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{20cm}-\frac{1}{u}=\frac{1}{10cm}$

$\Rightarrow \frac{1}{u}=\frac{1}{20cm}-\frac{1}{10cm}$

$\Rightarrow \frac{1}{u}=\frac{1-2}{20cm}=\frac{-1}{20cm}$
$\Rightarrow \frac{1}{u}=\frac{-1}{20cm}$
$\Rightarrow u=-20cm$

Therefore, the object is placed at $20cm$ away from the lens, opposite to the image side.

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

Putting values of $v$ and $u$:
Magnification, $m=\frac{h_2}{2cm}=\frac{20cm}{-20cm}$
$\Rightarrow \frac{h_2}{2cm}=-1$
$\Rightarrow h_2=-2cm$
Height of image is $2cm$, and the negative sign means the image is inverted.

Thus, the image is real, inverted and same as the size of the object.
The ray diagram shows the image formation.