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 = -0.1 m`, 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`
Required power of glasses, `P_g = P'_f - P_f = 50 - 60 = -10 D`
(ii) Power of accommodation `= 4D`
If `P_n` is power 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`.