Question

The near point of a hypermetropic eye is 1m. What is the nature and power of the lens required to
correct this defect
1) +3.0 D
2) + 4.0 D
3) -3.0 D
4) +2.0 D
​

Answer 1

Answer:

  3.)

Explanation:

Object distance, u = -25 cm

Image distance, v = -1 m = -100 m

Focal length, f

Using the lens formula,

1/f = 1/v - 1/u

here u = -25 cm

v = -100 cm

substituting & calculating for f, we get

f = + 33.3 cm = +0.33 m

Power P = 1/f = +3 D