An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.

(i) What is the nature of the image so formed ?

(ii) Calculate the distance of the image from the lens.

(iii) Calculate the magnification of the image.

Given u = 24 cm (-ve), f = 8 cm (+ve)

(i) Image formed is real, inverted and diminished.

(ii) Let v be the distance of image from optical centre. From the relation,

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

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{24} = \dfrac{1}{8} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{3 - 1}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{2}{24} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{12} \\[0.5em] \Rightarrow v = 12 \text{cm}$

Hence, the distance of the image from the lens = 12 cm

(iii) From relation, m = $\dfrac{v}{u}$

Substituting the values we get,

$m = \dfrac{12}{-24} \\[0.5em] \Rightarrow m = -0.5$

-ve sign of magnification tells us that the image is inverted.

Hence, magnification = -0.5