Question

S. We expect the normal eye to have a near point at 15 cm and a far point at infinity. A nearsighted patient comes for corrective contact lenses with a near point at 15 cm and a far point at 60 cm a) What type of lens will this patient need? Find ffor the correcting lens, and its strength in diopters. b) When wearing these lenses, where does the patient's near point appear to be? c) Recalculate the strength in diopters assuming the patient wants glasses (sitting 2.5 cm from the eye)

Answer 1

(a) To correct for this, we need to project an image at the far point

so,

image distance = - 60 cm

so,

1/f = 1/-60

f = -60 cm

P = 1/f

P = 1/0.6

P = - 1.667 D

This will be a diverging lens

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(b) Using lens equation

1/f = 1/p + 1/q

1/-60 = 1/p - 1/15

p = 20 cm

this is the new near point

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(C) if patient wants glasses

Here q = 60 - 2.5 = 57.5 cm

so,

f = - 57.5 cm

P = -1.739 D