Question

A concave lens forms a virtual image 0.4 times the size of the object. The distance between object and image is 5.3 cm. Find the focal length of the lens. Answer in units of cm.

Answer 1

Answer:

The focal length of the lens is -5.88 cm.

Explanation:

We know that magnification in case of lenses is given by

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

where

'v' is the position of image

'u' is the position of the object

Since it is given that m = 0.4 thus we can write

$\frac{v}{u}=0.4\\\\\therefore v=0.4u$

Now the distance between the object and the image in case of concave lens is given by

$u-v =5.3\\\\u-0.4u=5.3\\\\\therefore u=\frac{5.3}{(1-0.4)}=8.84cm\\\\v=0.4\times 8.84=3.54cm$

Now by lens formula the focal length of the lens is given by

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\\\frac{1}{f}=\frac{1}{-3.54}-\frac{1}{-8.84}\\\\\therefore f=-5.88cm$