A concave lens forms an erect image of 1/3rd the size of the object which is placed at a distance 30 cm in front of the lens. Find:

(a) The position of the image, and

(b) The focal length of the lens.

(a) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \qquad \bold{\text{(Equation 1)}}$

Given,

u = -30 cm

and

$m = \dfrac{I}{O} = \dfrac{1}{3} \qquad \bold{\text{(Equation 2)}}$

Substituting the values of u and Equation 2 in Equation 1, we get,

$\dfrac{v}{-30} = \dfrac{1}{3} \\[0.5em] v = -10 \\[0.5em]$

Therefore, the image is formed at 10 cm infront of the lens.

(b) As we know, the lens formula is —

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

Given,

u = -30 cm

v = -10 cm

The lens used is concave in nature.

Substituting the values in the formula, we get,

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

Therefore, the focal length is 15 cm (negative)