Question

a hypermetropic eye is corrected for near vision by lens of power 8/3 cal how is the near point of defective eye

Answer 1

Dear Student ,

Distance of the object for eye lens = Distance of normal near point, u = – 25 cm

Distance of the image from eye lens = Near point of hypermetropic eye v = ?

Power of lens = 8/3 = 100 /f

f = 37.5 cm

Now from the len's maker's equation we can write that ,

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{37.5}-\frac{1}{25}=\frac{1}{v} \\ \Rightarrow v=-75cm \end{gathered}$$
Regards