Question

the near point of hypermetrophic eye is 1 m what is power of lens required​

Answer 1

Answer:

The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.

Explanation:

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Answer 2

Answer:

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 :-

v1−u1=f1−1001−−251=f1f=3100=0.33mpower=f(inmetres)1=+3.0D