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If the aqueous humor of the eye has an index of refraction of 1.34 and the distance from the vertex of the cornea to the retina is 2.00 cm, what is the radius of curvature of the cornea for which distant objects will be focused on the retina? (For simplicity, assume all refraction occurs in the aqueous humor.)

5.07 \mathrm{mm}

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Rutgers, The State University of New Jersey

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Hope College

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'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'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's going to be Q times two minus and one over and two. No, I'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.

University of Wisconsin - Milwaukee