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University of Wisconsin - Milwaukee

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Problem 13 Easy Difficulty

An individual is nearsighted; his near point is 13.0 cm and his far point is 50.0 cm. (a) What lens power is needed to correct his nearsightedness? (b) When the lenses are in use, what is this person’s near point?

Answer

a. -2.00 \text { diopters }
b. 17.6 \mathrm{cm}

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Top Physics 103 Educators
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Video Transcript

So a person a near sighted person has his near point at 13 centimeter and Farpoint at 50 cent emitter. So let's find the power of the lens. Neither for this person. Okay, So to find the power, since the person is nearsighted, virtual and upright image should form at his fire point when he uses corrective lens to see a distant object. Okay, so that means P when P equals infinity, Q has to be at negative 50 centimeter. All right, so from the Lindsay craze in which is what over half? Because one of her p plus one over a cue, one of her pee is gonna be one of the infinity, which is zero and one of acute is going to be negative, one of our 50. So from here, you can see that actually, f is going to be negative 50 centimetres as well. So I'm just gonna write this down in meters. Your 0.5 meter. Okay. So from this information here, I can easily find the power the power require. Further corrective lens is going to be P equals one of her F, which is one over negative 0.5 that is going to give me two doctors negative to doctors. All right now to solve the part B to find out the person's near point when the leads are in use, we must know the fact that the object should form of virtual and upright emits at the near point of the eye without corrective bliss. Okay, so what that means is the image should be at the near point, meaning Q has to be called negative 13 centimeter in this case. So again, let's just use the lens equation. If you use the less equation that's going to give you. P equals Q times F over Q minus F and we know the F We calculated it, and this is going to be negative. 13 centimeter times, 50 cent emitter over negative 13 plus 50 Senator. Okay, so this will give us 17.6 centimeter. So this is going to be the object. A sense for that case

University of Wisconsin - Milwaukee
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Farnaz M.

Other Schools

Aspen F.

University of Sheffield