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.


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