a person affected by hypermetropia and his near point be 50cm and the power of lens given to the person
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Answered by
2
he should use convex lens
sampreethmadupalli:
i know but how to do the sum
then u should have np
pardon
i can't understand
Answered by
3
Here,
object distance, u = -25 cm
image distance, v = -50 cm
using lens formula,
1/v - 1/u = 1/f => -1/50 - (-1/25) = 1/f => 1/f = -1/50 + 1/25 => 1/f = (-1+2)/50 => 1/f = 1/50 => f =50 cm
so, focal length of the required lens = 50 cm = 0.5 m
Thus, power =1/f (in metres) = 1/0.5 = 10/5 = 2 D.
Therefore, the power of the required lens is 2 D.
object distance, u = -25 cm
image distance, v = -50 cm
using lens formula,
1/v - 1/u = 1/f => -1/50 - (-1/25) = 1/f => 1/f = -1/50 + 1/25 => 1/f = (-1+2)/50 => 1/f = 1/50 => f =50 cm
so, focal length of the required lens = 50 cm = 0.5 m
Thus, power =1/f (in metres) = 1/0.5 = 10/5 = 2 D.
Therefore, the power of the required lens is 2 D.
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