For those fortunate souls who do not need glasses, the lens of the eye adjusts i

Question

For those fortunate souls who do not need glasses, the lens of the eye adjusts its focal length in order to form a proper image on the retina. This typically means that very distant objects as well as objects as close as 25 cm can be seen clearly.
Many of us need corrective lenses since the lens in our eye cannot adjust sufficiently to produce a clear image over the full range object distances. This may be because the lens itself does not adjust well or because the eye is either longer or shorter than ‘normal’.
In the case of someone who is nearsighted (can see up close) the eye may only be able to see clearly items up to 50cm or 1m away (this would be the far point). In order to see something further away, a lens (either glasses or contacts) is used to produce a virtual image of a distant object at the person’s far point. Their eye can then accommodate the rest of the way and produce a clear image.
Suppose a person who has a far point of 66.3 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 66.3 cm from the lens?

Is the lens converging or diverging?

Correct: Diverging
Incorrect 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 54.3 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 54.3 cm from the lens?

Is the lens converging or diverging?

Correct: Converging
Incorrect Diverging

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

This discussion is closed.

Explanation / Answer

1. Using lens formula,

1/f = 1/v - 1/u ,here u = infinite ,v = -66.3cm

So f = -66.3cm , ..........(1)

Here f is negative so it is a diverging lens.......(2)

Power = 1/f = 1/0.663 = 1.508 ............(3)

Part B

Here u = -25 cm, v = -54.3cm

Using lens formula :

I/v - 1/u = 1/f

1/(-54.3) - 1/(-25) = 1/f

f = 46.33cm............(1)

f is positive so converging lens.........(2)

Power of lens P = 1/f = 1/0.4633 = 2.158 diopter.......(3)

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