The far point of a myopic eye is 20 cm . Find the power of his lens used.
3D
-3D
5D
-5D
Lens formula:
1v - 1u = 1f
u = - infinity
v = - 20 cm
1f = −120
f = -20 cm or −15 m
Power = 1f
P = -5D
The near point of a hypermetropic eye is 1 metre What is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye as 25cm)