Question

. A myopic person uses specs of power - 0.5D. What is the distance of far point of his eye​

Answer 1

Let far point of defective eye is v.

object distance , u = -∞

A myopic person uses specs of power, P = -0.5D

we know, power of lens , P = 1/focal length

so, -0.5D = 1/focal length

or, focal length = -1/0.5 = -2m

hence, focal length of specs , f = -2m

using lens maker formula,

1/v - 1/u = 1/f

or, 1/v - 1/-∞ = 1/-2m

or, v = -2m

hence, far point of his eye is 2m.

Answer 2

Answer:

A myopic person uses specs of power – 0.5D. What is the distance of far point of his eye?

(2m)