Question

The image of an object formed by a lens is of magnification-$1$ if the distance between the object and its image is $60$cm, what is the focal length of the lens? If the object is moved $20$cm towards the lens, where would the image be formed? State reason.

Answer 1

Given $m=-1,$

since $m=v/u=-1$ , -ive sign shows that $v$ is +ive i.e. image is formed on the other side of lens , and

$|v|=|u|$

now distance between object and image $=60cm$ ,

or $|u|+|v|=60$

or $|u|+|u|=60$

or $|u|=60/2=30cm$ or $u=-30cm$

and $|v|=30cm$ or $v=30cm$

now by lens formula ,

$f=uv/(u-v)=\frac{-30\times 30}{-30-30}=15cm$ (+ive focal length means it is a convex lens) .

Now if $u=-30+20=-10cm$ , as $u\left(10cm\right)<f\left(15cm\right)$ ,it means object lies between $F$ and $C$ , in this position of object the rays from object cannot meet at any point on the other side of lens ,which is also clear from the position of image found below ,

From lens equation $v=\frac{uf}{u+f}=\frac{-10\times 15}{-10+15}=-30cm$ ,

-ive sign shows that image will be formed on the same side of lens , where the object is placed .