Suppose a person who has a far point of 68.4 cm is trying to view a distant obje

Question

Suppose a person who has a far point of 68.4 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 68.4 cm from the lens?

Is the lens converging or diverging?

Diverging
Converging

Lenses are prescribed in terms of their refractive power, which is expressed in terms of diopters (see the text or your favorite search engine for the definition of a diopter). What is the refractive power of this lens in terms of diopters? (do not enter units.)

In the case of someone who is farsighted, the eye is not able to focus clearly on objects closer than a certain distance. This closest point on which a person’s eye can focus is called the near point.
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 57.3 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 57.3 cm from the lens?

Is the lens converging or diverging?

Converging
Diverging

What is the refractive power of this lens in terms of diopters? (do not enter units.)

Explanation / Answer

a. Image distance V = -68.4 cm, object distance u = infinity

From thin lens formula

1/f =1/u +1/v

1/f =1/-68.4 +1/infinity

f = -68.4 cm

this must be a Diverging lens as f is -ve
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b) Refractive power P=1/f in meters

P =-1/0.684

P= - 1/46 Diopters
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c)

Object distance u =25 cm

image distance v = - 57.3 cm

From thin lens formula

1/f =1/u + 1/v

1/f =-1/57.3 + 1/25

f = 44.23 cm

this must be converging lens as f is +ve

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d) Refractive power of lens P=1/f

P =1/0.4423

P = +2.26 Diopters

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