Question

An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of an object on a screen placed at a distance of 75 cm from the lens. Find- i) the focal length of the lens
ii) The magnification.

Answer 1

In this question we have been asked to calculate the focal length of the given system and the magnification of the lens. The distance of the image from the screen is given as 75 cm. Therefore, we will be using the basic lens formula to calculate the focal length of the lens and magnification formula to calculate the magnification.
Formula Used: -
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
$M=\dfrac{v}{u}$

Complete step-by-step solution
From the figure below, we can see that the distance of the lens from the screen where the image is obtained is 75 cm. Now it is given that the screen is 1 m away from the object as shown in the figure. Therefore, from the figure, we can say that distance of the object from the screen is -25 cm (using the sign conventions)

Therefore, we now have
$v=75;u=-25$
Now, to calculate focal length
From Lens formula we know
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
After substituting the values
We get,
$\dfrac{1}{f}=\dfrac{1}{75}-\dfrac{1}{(-25)}$
On solving
We get,
$\dfrac{1}{f}=\dfrac{1+3}{75}$
Therefore,
$f=18.75cm$
Now, for magnification we know
$M=\dfrac{v}{u}$
After substituting the values
We get,
$M=\dfrac{75}{-25}$
Therefore,
$M=-3$
Therefore, the focal length of the given lens is 18.75 cm and the magnification is -3.

Note: Magnification is the process of enlarging the apparent size of the given object. The physical size of the object is not changed. When the magnification of an object is less than 1 the object is said to be reduced in size. The distance from the center of the lens to the principal foci of a very thin lens is known as the focal length of the lens. For a converging lens, the focal length is positive.