A myopic adult has a far point at `0.1 m`. His power of accomodation is `4` diopters.
(i) What power lenses are required to see distant objects ?
(ii) What is his near point without glasses ?
(iii) What is his near point with glasses ? (Take the image distance from the lens of the eye to the retina to be 2 cm).

Explanation

(i) Let the power at the far point be `P_(f)` for the normal relaxed eye of an average person. The required power
`" " P_(f)=(1)/(f)=(1)/(0.1)+(1)/(0.02)=60D`
By the corrective lens the object distance at the far point is `oo`
The power required is
`" " P'_(f)=(1)/(f')=(1)/(oo)+(1)/(0.02)=50D`
So for eye+ lens system,
we have the sum of the eye and that of the glasses `P_(g)`
`:. " " P'_(f)=P_(f)+P_(g)`
`:." " P_(g)=-10D`
(ii) His power of accomodation is 4 D for the normal eye. Let power of the normal eye for near vision be `P_(n)`
Then, `" " 4=P_(n)-P_(f) or P_(n)=64D`
Let his near point be `x_(n)` , then
`" " (1)/(x_(n))+(1)/(0.02)=64 or (1)/(x_(n))+50=64`
`" " (1)/(x_(n))=14.`
`:. " " x_(n)=(1)/(14),0.07m`
(iii) With glasses `P'_(n)=P'P_(f)+4=54`
`" " 54=(1)/(x'_(n))+(1)/(0.02)=(1)/(x'_(n))+50`
`" " (1)/(x'_(n))=4`
`:. " " x'_(n)=(1)/(4)=0.25m`