Question

The near point of Hypermetropic eye is 100cm from the eye. What is the power

of the lens which he requires to read clearly a book held at 25cm from the eye? ​

Answer 1

Explanation:

the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence u=−25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v=−1m=−100 cm.

Using lens formula, we have :-

v

1

−

u

1

=

f

1

−100

1

−

−25

1

=

f

1 f= 3

100=0.33m power=

f(inmetres)

1 =+3.0D

Answer 2

Answer:

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