An object is placed at a distance of 20 cm in front of a concave lens of focal length 20 cm. Find —

(a) the position of the image, and

(b) the magnification of the image

(a) As we know, the lens formula is —

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

Given,

u = -20 cm

f = -20 cm

The lens used is concave, (as focal length is negative).

Substituting the values in the formula, we get,

$\dfrac{1}{v} – \dfrac{1}{-20} = \dfrac{1}{-20} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{20} = - \dfrac{1}{20} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{20} - \dfrac{1}{20} \\[0.5em] \dfrac{1}{v} = \dfrac{-1-1}{20} \\[0.5em] \dfrac{1}{v} = \dfrac{-2}{20} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{10} \\[0.5em] \Rightarrow v = -10 cm \\[0.5em]$

Therefore, position of image is 10 cm in front of lens on the same side of the object.

(b) As we know,

the formula for magnification of a lens is —

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

Given,

u = -20 cm

v = -10 cm

Substituting the values in the formula, we get,

$m = \dfrac{-10}{-20} \\[0.5em] m = 0.5 \\[0.5em]$

Therefore, the magnification is +0.5.