Question

(b) The near point of a hypermetropic eye is 1 m. Find the power of the
lens required to correct this defect. Assume that near point of the
normal eye is 25 cm​

Answer 1

Answer:

u = -25

v = -100 (1m = 100 cm )

1/f = 1/v - 1/u

1/f = 1/-100 -(1/-25)

= -1/100 + 1/25

= -1 +4 /100

= 3/100

f = 100 / 3

P = 100 / f

= 100 / (100/3)

= 100 × 3 /100

= 3D

power of lens should be 3D to correct the defect.

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