A person with a nearsighted eye has near and far points of11 cm and 35 cm,respec

Question

A person with a nearsighted eye has near and far points of11 cm and 35 cm,respectively.

(a) Assuming a lens is placed 2.0 cm from theeye, what power must the lens have to correct this condition?
P = 1 diopters

(b) Suppose that contact lenses placed directly on the cornea areused to correct the person's eye. What is the power of the lensrequired in this case, and what is the new near point? [Hint : Thecontact lens and the eyeglass lens require slightly differentpowers because they are at different distances from the eye.]
P = 2 diopters
p = 3 cm (a) Assuming a lens is placed 2.0 cm from theeye, what power must the lens have to correct this condition?
P = 1 diopters

(b) Suppose that contact lenses placed directly on the cornea areused to correct the person's eye. What is the power of the lensrequired in this case, and what is the new near point? [Hint : Thecontact lens and the eyeglass lens require slightly differentpowers because they are at different distances from the eye.]
P = 2 diopters
p = 3 cm

Explanation / Answer

a) since this eye can already focus on objects locatde at thenear point of a normal eye,no correction is needed for near objects.To correct the distant vision , a corrective lens(located 2cm from the eye) should form virtual images of very distant objects at 33cm in front of the lens.thus we have q =-32cm & P------> thus we have     1/f = 1/p + 1/q                = 0+1/-33cm or    f= -33cm refractive power is P = 1/-0.33m                                 = -3.03 diopters. b) A corrective lens in contact with the cornea should formvirtual image of very distant objects at the far point of theeye. therefore , we require that q = -35cm whenp------> thus we have 1/f = 1/p+1/q       = 0 +1/-35cm    or     f = -35cm we have refractive power as    P = 1/f = 1/-0.35m               = -2.857diopters when the contact lens (f = -35cm ) is in place, theobject distance which yields a virtual image at the near point ofthe eye (i.e q = -11cm) is given by      p = qf/q-f          =(-11cm)(-35cm) /-11cm+35cm          = 16 cm

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