A lens forms the image of an object placed at a distance 15 cm from it, at a distance 60 cm in front of it.

Find —

(i) the focal length,

(ii) the magnification, and

(iii) the nature of image.

As we know, the lens formula is —

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

Given,

v = -60 cm

So, u = -15 cm

Substituting the values in the formula, we get,

$\dfrac{1}{-60} – \dfrac{1}{-15} = \dfrac{1}{f} \\[0.5em] - \dfrac{1}{60} + \dfrac{1}{15} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{15} - \dfrac{1}{60} = \dfrac{1}{f} \\[0.5em] \dfrac{4-1}{60} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{60} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = \dfrac{60}{3} \\[0.5em] \Rightarrow f = 20cm \\[0.5em]$

Therefore, focal length of the lens is 20 cm.

ii) As we know,

the formula for magnification of a lens is —

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

v = -60 cm

u = -15 cm

Substituting the values in the formula, we get,

$m = \dfrac{- 60}{- 15} \\[0.5em] m = 4 \\[0.5em]$

Therefore, the magnification is 4.

(iii) The nature of the image is erect, virtual and magnified.