An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of the object on a screen placed at a distance of 75 cm from the lens.

Find —

(i) the focal length of lens, and

(ii) the magnification.

As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

Object lies at a distance 1.0 m from a screen.

v = 75 cm

So, u = - 25 cm

Substituting the values in the formula, we get,

$\dfrac{1}{75} – \dfrac{1}{-25} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{75} + \dfrac{1}{25} = \dfrac{1}{f} \\[0.5em] \dfrac{1+3}{75} = \dfrac{1}{f} \\[0.5em] \dfrac{4}{75} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = \dfrac{75}{4} \\[0.5em] \Rightarrow f = \text{18.75 cm} \\[0.5em]$

Therefore, focal length of the lens is 18.75cm.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

v = 75 cm

u = - 25 cm

Substituting the values in the formula, we get,

$m = \dfrac{75}{-25} \\[0.5em] m = -3 \\[0.5em]$

Therefore, the magnification is -3.