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A person with a nearsighted eye has near and far points of 16 cm and 25 cm, respectively. (a) Assuming a lens is placed 2.0 cm from the eye, what power must the lens have to correct this condition? (b) Suppose contact lenses placed directly on the cornea are used to correct the person’s eyesight. What is the power of the lens required in this case, and what is the new near point? Hint: The contact lens and the eyeglass lens require slightly different powers because they are at different distances from the eye.

a. -4.3 \text { diopters }

b. -4.0 \text { diopters }

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so a near sighted person has a near point at 16 centimeter and Farpoint at 25 cent. Vera. So in part A. If we want to find the correcting glance that are placed two centimeter in front of the persons, I then for a near sighted person to see the distant objects. That means when P is he called infinity, the image should form at fire Point, and it is virtually an upright. So the Farpoint image distance should be negative. 25 centimeter, right? But here's the thing, since the lenses place two centimeter in front of the eyes is actually going to be. The aim is distance is actually going to be 20 negative, 25 minus to subtract that and this is going to be negative. 23 centimeter. Okay, so the focal lens in this case when the object distances infantry is basically F because Q because negative 23 centimeter. All right, I'm just gonna write down this one in terms of meter. Is there a point to three meter? So from here, the power of the lens, which is Q. They're going to one of her focal length in matter, and this is going to be won over negative 0.23 This will give me negative 4.3 doctors. So we do not need correcting lens for the near point because the person is nearsighted and the normal near point of ah, regular normal near point off a person usual normal nor near point is 25 centimeter. But the person has the near point less than 25 centimeter. So we don't really need a correcting less for the your point for that person. Okay, so let's go to party now. If the contact lenses are used instead of my glasses, then your part B when the object is at infinity, when P equals infinity Q instead off negative. 23 centimeter is gonna be now, actually, negative 25 centimeter and the negative sign. I'm gonna mention that again. The negative sign is because the virtual a practice Mrs formed at No, the fire point of the person. Okay, so from here, the focal length again f is going to be called to Q, which is negative 25 sending meter and this is negative 0.25 meter. So the power of the contact list the person has to use has to be one of her f in meters, which is one over negative. 0.25 And this will give me negative four doctors. Okay, negative. Four adapters for this case.

University of Wisconsin - Milwaukee