Refraction Through a Lens — Question 13

As we know,

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

Given,

Power of the lens = -2.0 D

As the given power is negative hence, we can say that the lens used is concave in nature.

Substituting the value of power in formula we get,

$-\text{2.0} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.5m} \\[0.5em] \Rightarrow \text{focal length} = -\text{50cm} \\[0.5em]$

Therefore, the focal length is 50 cm and the lens used is concave in nature.