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The lens system of a person's eye has a power of 53.36 Diopters.Note that the distance from the eye's lens to the retina is 2.0 cm.For this person's ...

Question

The lens system of a person's eye has a power of 53.36 Diopters.Note that the distance from the eye's lens to the retina is 2.0 cm.For this person's eye what distance from the eye must an object beto be in sharp focus?

The lens system of a person's eye has a power of 53.36 Diopters. Note that the distance from the eye's lens to the retina is 2.0 cm. For this person's eye what distance from the eye must an object be to be in sharp focus?



Answers

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.

In this problem, given that B is equal to two centimeters which is equal to 0.2 me. That U. Is equal to minus 10 centimeter, which is equal to minus 0.1. Meted now going forward using the land, formula one by B minus one by U. Is equal to one by yep. So putting the value in this expression, I can write one by your physical, it won by 0.0 to minus one by minus 10.1. We turn simplification. I can write the value of one by F is equal to 50 plus 10 which is equal to 60 m. So power is equal to one by F which is equal to 60 director. As the answer. Now here the value of U is equal to minus 100 centimeter which is equal to minus one m. He is equal to two centimeter which is equal to 0.2 m. I will use the formula one by B minus one by U. Is equal to one by F. On further simplification, I can write the value of one by F is equal to one by one year or two minus one by minus one. Which on simplification I can write the value of one by F is equal to 50 plus one, which is equal to 51 power of the lens, is equal to one by F, which is equal to 51 director, which is equal to 51 director. So they arrange it from plus 60 director to plus 51 director.

So in this question, the distance from the lens from the lens to the retina at the back of the eye, it is equal to 2.0 centimeter. So this will be the major distance because he misses formed at the regina. Okay, if the light is to focus on the retina, so for the part we have to determine the focal length of the lens. When viewing a distant object. That is objective distance is taken as infinity. So from the length formula we can write that one by D. O. Plus one by D. I. It is equal to one by F. So substituting the value, so one by deal which is infinity plus one by D. I. Which is 2.0 centimeter, it is equal to one by F. So from here focal length of the lens comes out to be 2.0 cm. So this become the answer for the part of the problem. Okay, now, solving for the part B in which we have to determine the focal length of the lens. Again, When viewing an object which is at a distance D. O. It is equal to 25 cm away from the front of the eyes in front of the it is given. So these relations will be positive. Okay, now, applying this formula again, so we can write that one by. Do you? It is equal to this 25 cm plus one by D. Which is 2.0 centimeter. This is equal to one by F. So from here after solving focal length will be equal to 1.9 centimeter. So this becomes the answer for this problem. Okay, thank you.

Power is given by one of s because why would be just a few for the camera? The object distances infinity on. He made his form at the distance from the lens. Kiddo, Whatever we consider four and zero want a we don't so that becomes 56 Maker in verse. It was 56 about that. Now, In the second case, if we choose power at the object distance to be 0.2 m on the immense distance K's as it applies to what a meter. Then doing the same competition. We will find the power to be 61 doctor on. In the third case, if you have the object at the same distance, however, your images Marty one. So that's a negative sign on its one middle of a. Then, in this particular case, even the same calculation you find the power before

The asserted parson needs 1 kg now for the object. That's that infinity. We would have to find the focal in because one over do you. That's one of our D I. And worse object is listed in unity. So that becomes equal to D I. It was negative. 80 centimeter. So focus and become negative. 80 centimeter and power becomes one over. If it was one over negative 0.8 doctors, it was 1.255 30 the negative.


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