The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 20 cm
behind the eye-lens, what is the range of the power of the eye-lens?
Given data -
Near point of the child ,u = 10 cm = 0.1
Far point of the child = 100 cm
Distance of the retina from the eye lens, v = 2cm = 0.02m
Using lens formula –
Substituting the values-
Power of the lens when the near sight is 10cm,
When, the near point of the child is 100cm = 1m, u = -1m
Using lens formula –
Substituting the values-
Power of the lens when near point is 1m-
So, for the near point and the far point of a child to be10 cm and 100 cm, eye lens
power should lie between+ 60 D to + 51 D.