Question

Mike can see clearly only the objects situated in the range of 50 cm to 300 cm. The optometrist prescribed him a lens to increase the maximum distance of vision to infinity. Upon using the prescribed lens, he discovered that the near point of his vision has shifted from 50 cm to a distance ‘d’. What is the value of ‘d’?

Answer 1

Step 1: Calculating the Power of the prescribed lens
Given: v = -300 cm, u = ∞
1/v - 1/u = 1/f
1/(-300) - 1/∞ = 1/f
1/f = -1/300 cm-1 = -⅓ m-1
P = 1/f = -1/3 = -0.33 D

Step 2: Calculating the value of ‘d’
f = -300 cm ; v’ = -50 cm; u’ = d
1/v’ - 1/d = 1/f
1/(-50) - 1/d = 1/(-300)
1/d = -1/60
d = 60 cm