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University of Wisconsin - Milwaukee

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Problem 15 Medium Difficulty

A particular patient’s eyes are unable to focus on objects closer than 35.0 cm and corrective lenses are to be prescribed so that the patient can focus on objects 20.0 cm from their eyes. (a) Is the patient nearsighted or farsighted? (b) If contact lenses are to be prescribed, determine the required lens power. (c) If eyeglasses are to be prescribed instead and the distance between the eyes and the lenses is 2.00 cm, determine the power of the required corrective lenses. (d) Are the required lenses converging or diverging?

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Video Transcript

In this problem, a patient can focus in the objects that are closer than 35 centimeters, which means he has a near point at 35 centimeters. So which also means the person is far sighted. Okay, The person is far sighted. And the reason is because he cannot focus on the objects that are closer than 35 centimeters. All right, so now if the corrective lenses are prescribed so that the patient can actually see the objects at 20 centimeter from the I, then then virtual a pride image of an object that is at Pecos 20 centimeter must be formed at the Miss distance of the the near point of deer person. Okay, so that means thank you. Has to be negative. 35 centimeter. So I'm just gonna write this down in meters, zero point two meter. And this one is that negative? 0.35 meter. All right, so from here it is pretty straightforward to find the power of the lens. So all we have to do is use the lens equation. So from here, if we use the less equation key, the power is going to be won over F, which is equal to one of her small P plus one of her small Q. And that is going to be 1/0 0.2 minus one of her 0.35 And this will give us positive 2.14 adapters. All right, so now instead of context, if eyeglasses are to be prescribed and the distance between the eye and the lenses is two centimeter, then what do we do? Let's find out. So the new object distance in this case from the corrective less would be P. He calls 20 minus two centimeter. It calls 18 cent emitter because I'm gonna write down in meters, which is 0.18 meter. Remember, I'm subtracting two from 20 here because there's a distance to centimeter between the eye and the contact lens. So similarly, for the team is distance. The new of the the Emma's distance would have to be Q. It call 35 minus two centimeter, which is going to be negative 0.33 meter. All right, so now, in this case, the new power would be P for the glasses would be one of her F, which is one of her P 1st 1 of acute. And this is going to be cool too. 1/0 0.18 minus one over sirrah 0.33 That is going to give us a positive 2.53 doctors. So the less is still going to be the converging, less system. Power is positive. Positive 2.53 Right, So the lens is converting.

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

Cornell University

Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Meghan M.

McMaster University