Suppose a person who has a far point of 79.2 cm is trying to view a distant obje

Question

Suppose a person who has a far point of 79.2 cm is trying to view a distant object. What is the focal length (with correct sign) of a lens that would take a distant object and make an image on the same side of the lens as the object a distance 79.2 cm from the lens? Lenses are prescribed in terms of their refractive power, which is expressed in terms of diopters. What is the refractive power of this lens in terms of diopters? In this situation the corrective lens is used to make an object closer than the near point produce an image further away from the lens at the near point.
Suppose a person who has a near point of 37.3 cm is trying to view a book at a distance of 25.0 cm. What is the focal length (with correct sign) of a lens that would take the book and make an image on the same side of the lens as the book a distance 37.3 cm from the lens? What is the refractive power of this lens in terms of diopters?

Explanation / Answer

To correct nearsightedness the object distance is infinity and the image is at -79.2cm

So from 1/f = 1/do + 1/di we have f = di = -79.2cm which is -0.792m

So the power is 1/-0.792 = -1.262 diopter
and the lens is diverging

2) We have do = 25cm and di = -37.3cm

So 1/f = 1/do + 1/di = 1/25 + 1/(-37.5) = 1.118x10^-2

So f = 1/1.118x10^-2 = 89.44cm

So power = 1/0.8944m = +1.11 Diopters (converging lens)

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