Refraction Through a Lens — Question 2

(a) As the lens used is a concave lens, so the image formed will be erect and diminished. It will be virtual in nature as the image is formed on the same side as the object.

(b) As we know, the lens formula is —

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

Given

u = -20 cm

v = – 10 cm

Substituting the values in the formula we get,

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

Therefore, focal length of the lens = 20 cm (negative)