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=-0.1 m
Image distance, v=+0.02 m
The power, Pf=1f=1v-1u=10.02+10.1=60 D
By the corrective lens, the object distance at the far point is ∞.
The required power is,
P'f=1f'=1∞+10.02=50 D
The power of the glasses,
Pg=50 D-60 D=-10 D
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-Pf or Pn=64 D
Let his near point be xn, then,
1xn+10.02=64 or 1xn+50=64
xn=114
∴ xn=114=0.07 m
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= 0.25 m
© 2025 GoodEd Technologies Pvt. Ltd.