Question

The near point of hypermetropia eye is 1m . What is the power of lens required to correct this defect ? Assume near point of eye is 25 cm

Answer 1

Answer:

Answer:− C option

In 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

Explanation:

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