Solved Sample Paper 4 — Question 4

(a) The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power. It can be defined as a reciprocal of focal length:

$\text{Power} = \dfrac{1}{\text{focal length}}$

(b) As the given power is negative hence, we can say that the lens used is concave in nature and the child is suffering from myopia or short sightedness (i.e., unable to see the far objects distinctly).

(c) As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = -2.5 D

Substituting the value of power in formula we get,

$-\text{2.5} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2.5} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.4 m} \\[0.5em] \Rightarrow \text{focal length} = -\text{40 cm} \\[0.5em]$

∴ Focal length = - 40 cm.