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Question

The near point of a hypermetropic eye is 1 metre. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.


Solution

Here, u = -25 cm = -0.25 m
v = -1 m
Now, P=1f=1v1u=1110.25 =1+4=+3D

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