The far point of a myopic eye is 60 cm. What is the nature and power of the lens required to correct the problem
The focal length required is
1/f = 1/v - 1/u
here u is taken as infinity and v is -60 cm because the person can see an object up to 60 cm. So a distant object is considered such that the image is formed at 60 cm
Therefore
1/f=1/(-60) - 0
f=-60 cm=-0.6 m
Power = 1/f = 1/(- 0.6)= -1.667 Diopter.