Thank you for reporting, we will resolve it shortly
Q.
The nearer point of hypermetropic eye is $40\, cm$. The lens to be used for its correction should have the power
Ray Optics and Optical Instruments
Solution:
Near point of a hypermetropic eye $=40 \,cm$
Near point of a normal eye $=25\, cm$
Object distance, $u =-25\, cm$
Image distance, $v =-40\, cm$
Using lens formula,
$ \frac{1}{ f }=\frac{1}{ v }-\frac{1}{ u }=\frac{1}{-40}+\frac{1}{25} $
$ \frac{1}{ f }=\frac{-5+8}{200}=\frac{3}{200 cm }=\frac{300}{200 m } $
$ P =\frac{1}{ f ( in m )}=\frac{300}{200}=\frac{3}{2} $
$ P=1.5 \,D $