Question

For a normal eye the near point is 25 cm from the eye lens. The distance between eye lens and retina of the eye is 2.5 cm. what is power of eye lens?
options are:-
(a)4
(b) 44
(c) 40
(d) -44

Answer 1

$$\begin{gathered} nearpointistheobjectdis\tan ceu=-25cm \\ dis\tan cebetweenretinaanseyelensi.e.objectdis\tan ce=2.5cm \\ focallengthoflensisgivenby: \\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{f}=\frac{1}{2.5}+\frac{1}{25} \\ \frac{1}{f}=\frac{11}{25} \\ f=\frac{25}{11}cm=\frac{25}{1100}m \\ powerofthelens=\frac{1}{f}=\frac{1100}{25}=44D \end{gathered}$$