A person using spectacles having power `+2.5m^(-1)` can see the objects distinctly at a distance of 25 cm . What is the near point for the person ? What type of defect of vision does the eye have ?

Explanation

Power of spectacles , P =+2.5`m^(-1)`. If f be the focal length of the lens of the spectacles, then
P=`(100)/f or, 2.5 =(100)/f or, f =(100)/(2.5) =40 cm `
Suppose, the distance of the near point of the defective eye =x cm
The lens will form the image of the object situated at a distance of 25 cm at a distance x cm .
Here u=-25 cm , v=-x cm and f =40 cm .
From the equation `1/v-1/u=1/f` we have
`1/(-x)-1/(-25)=1/(40) or, 1/x =1/(25)-1/(40)=3/(200) or, x =(200)/3 =66.67 cm`
This is the distance of the near point for the person i.e , the near point has been shifted away from normal distance (25 cm ). so the defect of the eye is long sight .