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) Distance of far point, `u = - 0.1m`, distance of image, `v = 2 cm = 0.02 m`.
`:.` Power of myopic eye, `P_(f) = (1)/(f) = (1)/(v) - (1)/(u) = (1)/(0.02) + (1)/(0.1) = 50 + 10 = 60 D`
With the corrective lens, far point shifts to `oo`. `:.` Power required, `P'_(f) = (1)/(0.02) - (1)/(oo) = 50 D`
(ii) Power of accomodation `= 4D`
If `P_(n)` is power of normal eye for near vision, then `4 = P_(n) - P_(f) = P_(n) - 60 or P_(n) = 64 D`.
If `x_(n)` is near point without glasses, then `(1)/(x_(n)) + (1)/(0.02) = 64 or (1)/(x_(n)) = 14, x_(n) = (1)/(14)m = 0.07 m`
(iii) With glasses, `P'_(n) = P'_(f) + 4 = 50 + 4 = 54`
`:. (1)/(x'_(n)) + (1)/(0.02) = (1)/(x'_(n)) = 54 (1)/(x'_(n)) = 54 - 50 = 4`
`x'_(n) = (1)/(4)m = 0.25 m`