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 ? (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

`rArr` (i) Required power for normal relaxed eye at far piont,
`P_f = 1/f= 1/f_1 + 1/f_2`
`therefore P_f = (1)/(0.1) + (1)/(0.02) = (0.02 + 0.1)/(0.002) = (0.12)/(0.002)`
`therefore P_f = 60 D`
`rArr` Now required power for shifting far point to infinity (using corrective lens)
`P._(f) = (1)/(f.) = (1)/(oo) + (1)/(0.02) = 0 +(1)/(0.02)`
`therefore P.f = 50 D`
For the combined system of relaxed eye and spectacles,
`P._(f) = P_(f) + P_(g)`
`therefore 50 = 60 + P_g`
`therefore P_g = -10 D`
(ii) Power of normal eye P = 4D
Now, if power eye P = 4D
Now, if power of normal eye of given person for near vision is `P_n` then ,
`P = P_n - P_g`
`therefore 4 = P_n -60`
`therefore P_n = 64 D`
`rArr` Suppose distance of near point without spectacles is `x_n`
`therefore 1/(x_n) + (1)/(0.02) = 64`
`therefore (1)/(x_n) + 50 = 64`
`therefore 1/(x_n) = 14`
`therefore x_n = (1)/(14)`
`therefore x_n = 0.07 m `
(iii) With spectacles,
`P._n = P._f +P`
`54 = (1)/(x._n) +(1)/(0.02) = (1)/(x._n) + 50`
`therefore (1)/(x._n) = 4`
`therefore x._(n) = 1/4 = 0.25 m `