Q. 20

# 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?

Answer :

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.

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