Physics, asked by Dhruv413, 11 months ago

A hypermetropia eye is corrected for near vision by using a lens of power 8/3 d. Calculate how far is the near point of defective eye?draw ray diagram for corrected vision

Answers

Answered by Abyze159

3

Answer:

37.5cm

Explanation:

put 8/3=100/f

i.e. f=300/8=75/2

=37.5

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