(a) At what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at a distance 30 cm behind the lens?

(b) What is the focal length of the lens in part (a)?

(a) As we know,

The formula for magnification of a lens is —

$m = \dfrac{\text{length of image (I)}}{\text{length of object (O)}} = \dfrac{v}{u} \\[0.5em]$

where,

v = image distance

u = object distance.

Given,

Height of a candle, O = 3 cm

Height of the image of candle, I = 6 cm

Image distance, v = 30 cm

Substituting the values in the formula we get,

$\dfrac{6}{3} = \dfrac{30}{u} \\[0.5em] u = \dfrac{30 \times 3}{6} \\[0.5em] u = 5 \times 3 \\[0.5em] \Rightarrow u = 15$

Therefore, object distance is equal to 15 cm.

(b) As we know, lens formula is —

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

Substituting the values in the formula we get,

$\dfrac{1}{30} – \dfrac{1}{-15} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{30} + \dfrac{1}{15} = \dfrac{1}{f} \\[0.5em] \dfrac{2 + 1}{30} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{30} = \dfrac{1}{f} \\[0.5em] f = \dfrac{30}{3} \\[0.5em] \Rightarrow f = 10$

Hence, the focal length of the lens, f = 10 cm.