Due in 13 hours, 52 minutes For those fortunate souls who do not need glasses, t

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Due in 13 hours, 52 minutes 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 70.1 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 70.1 cm from the lens? Submit Answer Tries 0/10 Is the lens converging or diverging? Diverging Converging Submit Answer Tries 0/10 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.) Submit Answar Tries o/10 In the case of someone who is farsighted, the cye is not able to focus clearly on objects closer than a certain distance. This closest point on which a person's cye 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 53.0 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 53.0 cm from the lens? Submit Answer Tries 0/10 Is the lens converging or diverging? Diverging Converging Submit Ans Tries 0/10 What is the refractive power of this lens in terms of diopters? (do not enter units.) Submit Ans Tries 0/10

Explanation / Answer

(a)

Person has a far point of 70.1cm.

The lens used should be able to form a virtual image of far distanced objects at 70.1cm, which then acts as the object for the lens of eye. The eye can comfortably adjust to view the object which is in its far point range.

For best puposes the lens used should form an image at 70.1cm for an object at infinity.

So: 1/v - 1/u = 1/f (Lens makers formula)

with u=infinity v = -70.1cm ( negative as the image formed is in front of eye but the direction of light is towards the eye)

Substituing we get f = -70.1cm

the focal length is negative indicating a concave lens, i.e. a diverging lens

Power in dioptres is given by : P = 1/f (f in metres)

P = 100/(-70.1) = -1.426 dioptres

(b)

Person has a near point of 53.0cm.

The lens used should be able to form a image of objects closer than 53cm at far distances, which then acts as the object for the lens of eye. The eye can comfortably adjust to view the object when it is outside its near point.

A regular eye can see close objects upto 25cm. So the lens in best condition should form the image at 53cm for an object at 25cm.

1/v - 1/u = 1/f (Lens makers formula)

with u= -25cm v = -53cm ( negative as the image formed is in front of eye but the direction of light is towards the eye)

Substituting and calculating : we get f = +47.32 cm

the focal length is positive indicating a convex lens, i.e. a converging lens

P = 100/(+47.32) = +2.113 dioptres

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