Question

The object is at distance 45 cm from the screen. With the lens, we obtain a small image of the object on the screen. By moving the lens. We receive a different image on the screen, whose size is 4 times greater than the first. What is the focal length of the lens (in cm)?

Answer 1

Let, the object distance be $u$ and the image distance be $v$.

Since the screen is $45$ cm away from the object.

$u+v=45$

Since the image is formed behind the lens, the image is real.

We know,

$m=-\frac{v}{u}$

$\Rightarrow -m=\frac{v}{u}$

$\Rightarrow -4=\frac{v}{u}$

$\Rightarrow v=-4u$

Putting this value in the above equation,

$u+v=45$

$\Rightarrow u-4u=45$

$\Rightarrow -3u=45$

$\Rightarrow u=-15$

$\therefore v=-4(-15)$

$=60cm$

By Lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{60}-\frac{1}{-15}$

$\frac{1}{f}=\frac{1}{60}+\frac{1}{15}$

$\frac{1}{f}=\frac{1+4}{60}=\frac{5}{60}$

$\frac{1}{f}=\frac{1}{12}$

$f=12cm$

$\therefore$ The focal length of the lens is $12cm$.