Question

9) Focal length of a lens is 20 cm, find its power. What eye defect is it used
to correct?
0.05 D, hypermetropia
0.05 D, myopia
0.5 D, myopia
0 0.5 D, hypermetropia​

Answer 1

Answer:

5D will be the answer

Explanation:

P = 100/f (f in centimetres)

P = 100/20 = 5 D

Answer 2

Answer:-

5D will be the answer

Explanation:

P = 100/f (f in centimetres)

P = 100/20 = 5 D