the-lens-in-a-normal-human-eye-with-aqueous-humor-on-one-side-and-vitreous-humor-on-the-other-side-h

Question

The lens in a normal human eye, with aqueous humor
on one side and vitreous humor on the other side, has a refractive
power of 15 diopters. Suppose a lens is removed from an eye and
surrounded by air. In this case, is its refractive power greater than,
less than, or equal to 15 diopters? Explain.

Nathan Silvano · Accepted Answer

Okay, so in this problem, we have the lens of a human eye. Let&#x27;s represent with this straight line black straight line and the lens of the human rights is surrounded in one side by aqua US humor and on the other side by a Viet vitreous humor. No. Suppose that actually this system has a refractive power. Let&#x27;s go reflective power A were effective power off, 15 day up nurse. Then we take the lands off and surround the lens in the air. So now the lands are surrounded by air and the problem wants to know if the reflective power off the second situation let&#x27;s cover effective power. Bi. This one reflective power eight is going to be last equal or greater. And in the first interior, so think of this when we put the land surrounding in the air. The difference in index of refraction between the lens and its surroundings is always greater then, when it&#x27;s surrounded by the aqua us and the vitreous humor. So because of the index of refraction, of the air is less off this off his humors, they&#x27;re effective. Power needs to be greater, so we can say that they&#x27;re effective power off the lens in the air is greater, then 15 die afters, and we just need to know that the index of refraction off the air this is smaller than in the other in the other humors. Therefore, the refractive index is going to be greater, and that&#x27;s

Netra Sharma · Answer

so we have. The reflective index of the Equus humor is 1.34 The distance from the verdicts of the cornea to Regina, which is also the immense distance, is two centimeter now for the distant objects to be able to focus Andrew Tina, which is if we want the object, this tends to be infinity, and they missed a sense to be to send a reader. And assuming that all the refraction occurs in Echo&#x27;s humor, we actually use the relation for the refraction through this very cool surface on DA. The religion for this vehicle surface is perfection through the spherical surface is and one over P plus and two over Q because and two minus and one over. Yeah, so here and one is the refractive index off air. Okay, which is one and and two is actually end, which is the effective index, of course, former, which is 1.34 and our is the radius of curvature of the the entire I. All right, so we want to find out the radius of curvature here. Let&#x27;s go to a second piece for that one. So from the relation we have, we can actually rewrite it and right in terms of radius of curvature, and it&#x27;s going to be Q times two minus and one over and two. No, I&#x27;m just gonna plug in the value for everything here. Q is two centimeter and two is 1.34 and then one is air, which is one over and two is 1.34 again. So from here we can actually get 0.507 centimeter or 5.7 millimeter. That is the radio radius of curvature of the eye.

Mayukh Banik · Answer

so we&#x27;re going to use Does miss, though at the lens I interface and salt for the angle. So if this is the victorious humor, then the angle off refraction in the I. Instead I we had which we saw to be 24.6 degree. Now 90 minus state. I will be the angle with respect to the surface, which is 65.4 degrees.

Sean Dougherty · Answer

imagine that. We&#x27;re gonna look at this circle right here and looking at the circle from the side. So that&#x27;s what looks like a novel. Um, we&#x27;re gonna look at it from a point over here. Now we want to be able to resolve the point on the very bottom of the circle and a point on the very top. So you know that it&#x27;s a circle and not just a fuzzy dot so we can draw this kind of triangle. We know that this circle is going to be our near point 4 25 centimeters and we want to find that minimum separation, and then we&#x27;ll circle diameter that will. I was to see a circle and not just again one point. Okay, well, this angle here in this triangle is that fatum in or Alfa we can express fate of man has 1.22 lambda over de We also can express fatum in, has an approximation for 10. Feed him in because the small angle tan Federman is just fada, which is equal to the opposite of this triangle from fate of man, which is demon over adjacent. Just l tend to be clear. Ellis this one so we can express D men as, um 1.22 times our wavelength in the eye, which we need to adjust for. So start the wavelength of hair divided by an excellent fraction for a I, uh, him. In times this l over d the diameter of the villains and I So we plug in our numbers. Demon is 1.22 times 6.0, whoops. Times 600 dumped in negative nine meters Rex one of work in meters, uh, over 1.33 times L, which is 25 centimeters or 0.25 meters and then our lens diameter his two millimeters or 0.2 meters. That minimum diameter of a circle. The near point will be 6.9 times 10 to the negative five meters or 0.69 millimeters.

Mayukh Banik · Answer

in? Sure. Och user. Then he&#x27;s implant dead to trade my obeah. And it was held, the person read without glasses.

Farhanul Hasan · Answer

All right. So the cornea is treated as a thin lens here, and so that then leads equation applies. One of her focal length is equal to refractive index minus one times one of radius of curvature of fun and minus one. Overuse. Um, coach of the back it And it is this our two that we want here in part eso let&#x27;s we&#x27;re angels equation and write one of our one writes or one of our two is equal to one over f times one over and minus one. That&#x27;s just, um, pulled to the left hand side on dso One of our two is just one of her. Are one minus this whole thing of one of our aft times one over and minus one. Okay, on DSO. One of our one is five millimeters less of directing staff. Trip to millimeters. Ah f is 1.8 centimeters. Ah, so that will be 1.8 centimeters are about 18 millimetres. 18 millimeters. Oops. Okay, so that&#x27;s 18 millimeters, and that&#x27;s and er multiplied by one over one point 38 minus 11 country. It was a given refractive index. That&#x27;s one of our two and So this implies that are too Take a reciprocal. What you&#x27;ve got get is 18.6 millimeters or 1.86 centimeters. Part B wants the image distance and so he is one over f equals one of us, plus one over X prime and one of us plus one arrest prime So hopes Assassins as prime there on dso one arrest as primarily once So could be right. This when I were f minus one over R s and half again is eyes one little strain two centimeters from 20.8 centimeters. That&#x27;s F and s image distances 20 Object distance 25 And so we get that as prime is 1.9 centimeters. Finally for part. See Ah, we first configure out magnification shows negative image distance over object distance. So that&#x27;s negative. 1.9 over 25. That&#x27;s negative. 250.76 This means that the images, um inverted lips on inverted, um, and then because it&#x27;s negative and then em is also God, em is also Ah, negative. Whoops. Positive. Why prime over why My prime is what we want. The image height. So that&#x27;s m times. Why. So that&#x27;s negative 0.76 times eight millimeters. And so this comes out to be negative 80.61 millimeters. So it&#x27;s a D magnified image and then ask prima&#x27;s positive. Let&#x27;s be found in part B. So this is ahh, real image.

Problem

Predict/Explain Treating Cataracts When the lens …

Problem 82 Medium Difficulty

The lens in a normal human eye, with aqueous humor
on one side and vitreous humor on the other side, has a refractive
power of 15 diopters. Suppose a lens is removed from an eye and
surrounded by air. In this case, is its refractive power greater than,
less than, or equal to 15 diopters? Explain.

Answer

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