A myopic adult has a far point at 0.1 m. His power of accommodation is 4 D.
(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.)

Hint: The image formed by the lens acts as an object for the eye-lens.

Step 1: Find the power of the glasses.

(i) For the myopic person,

Object distance, u=-0.1m

Image distance, v=+0.02m

The power, Pf=1f=1v-1u=10.02+10.1=60D
By the corrective lens, the object distance at the far point is .
The required power is,
                            P'f=1f'=1+10.02=50D
The power of the glasses,
Pg=50D-60D=-10D

Step 2: Find the near-point without glasses.

(ii) His power of accommodation is 4 D for the normal eye. Let the power of the normal eye for near vision be Pn.
Then,                          4=Pn-PforPn=64D
Let his near point be xn, then,
                                           1xn+10.02=64or1xn+50=64
                                           xn=114  
                                        xn=114=0.07m
Step 2: Find the near-point with glasses.
(iii) With glasses P'n=P'f+4=54
                                             54=1x'n+10.02=1x'n+50
1x'n=4

                                        x'n=14=0.25m