A simple and relatively inexpensive microscope eyepiece is the...

Question

A simple and relatively inexpensive microscope eyepiece is the Ramsden eyepiece shown in FIGURE P35.40. Two plano-convex lenses have their curved surfaces facing each other, which a more advanced analysis shows is the orientation that minimizes spherical aberration. That same analysis finds that chromatic aberration is minimized with lens spacing L = 1/2 (f₁ + f₂). Your task is to design a 10x Ramsden eyepiece in which the first lens has a focal length of 30 mm. What are (a) the focal length and (b) the spacing of the second lens?

Accepted Answer

Welcome back. Everyone in this problem. The Huygens I piece follows the principle of minimizing spherical aberration by arranging two plano convex lenses with their curved surfaces facing each other. An analysis reveals that chromatic aberration is minimized for the Huygens I piece. When the spacing between the lenses denoted as L equals half the sum of their focal lengths. Let's shift our focus to the design of a Huygens IPI aiming to achieve a 12 times magnification. Given that the first lens of the eye piece has a focal length of 35 millimeters, we want to calculate one, the focal length of the second lens in the eye piece and the two, the spacing between the 1st and 2nd lens in the hurricanes ice. Now, let's start by trying to find the focal length of the second lens in the hurricanes eyepiece. What do we know? Well, recall that if we're talking about an eyepiece, the magnification of an eyepiece, OK. Equals for relaxed viewing, that is equals 25 centimeters divided by the focal length of the IPs that tells us then that the focal length for I piece is going to be equal to 25 centimeters divided by the magnification of our IP, which in this case, we're trying to achieve a 12 times magnification. So that's 25 centimeters divided by 12 and that equals 2.083 centimeters. OK. So that would have been the focal length for our IP. We figured out the first part of our problem. Next, let me put that in blue here. We'd like to find the spacing between the 1st and 2nd lenses in the Huygens eyepiece. Now, what do we know about the spacing? Well, recall in our problem, it says that our chromatic aberration is minimized when the spacing between the lenses equals half the sum of their focal length. And we're even given a formula here for that length. So therefore, our spacing o would be half of the focal lens of both eyes of both lenses. OK. Now, the thing is here, we know the focal length of our first lens is our first a piece is 35 millimeters, but we don't know the focal length of our second length. So that's what we need to figure out. If we can figure out what that focal length is, then we should be able, let's designate it rather as F two, then we should be able to figure out the spacing needed. Now, what else do we know? Well, we also know that the effective focal length of our real lenses, which would be the focal length of our ips here So one divided by the effective focal length would be equal to the reciprocal of the first focal length plus the reciprocal of the second focal length minus or spacing between the lenses divided by F one multiplied by F two. Now we know the space, the spacing between our lenses here. So we can substitute that formula into the formula for the effective focal lens to be the reciprocal of F one plus the reciprocal of F two minus substitution formula half of F one plus F two. So that would be F one, F two divided by two multiplied by F one F two. Now here, since we know that our effective focal length is the same thing as the focal length of our eyepiece. And we know what the focal length of our first lens is. We should be able to solve the for F two. And if we solve for F two, then we should be able to tell the separation between the lenses. So let's go ahead and do that. So here, if we first of all try to get one fraction on the right hand side of our equation, then one divided by effective focal lengths, we're gonna do a bit of math here is going to be equal to, well notice all three denominators have F one F two in common or two multiplied by F one F two. So F one into two multiplied by F one F two equals two F two. OK. F two into two, multiplied by F one, F two equals two, F one. And for F one, F two divided by two F one F two, we'd still get back our, our numerator. And if we're subtracting the expression, that means we're subtracting F one by the distributive law and subtracting F two. And all of that is being divided by two F one F two. Now, we can simplify here two F two minus F two. That's just F 22, F one minus F one. That's just F one. And that is going to be divided by two F one F two. Now notice here both terms in our numerator have something in common with the denominator. So we can split them up and south because we can rewrite this then as F 2/2, F two, F one plus F 1/2 F two, F one. OK. I don't know if two cancers F two in our first fraction to give us uh to give us uh 1/1, divided by two F one and F one cancels F one in the second fraction to give us one divided by two F two. Now we have F two by itself in a term. So we're closer to solving for F two because now this tells us that one divided by two F two is going to be equal to one divided by our effective focal length minus one, divided by two F one. And if we reciprocate four F two, that means then that two, F two, OK is going to be equal to the reciprocal of difference. So that's one divided by effective focal length minus one multiplied by two F one to the power of negative one. So therefore, F two, we can take over if we divide both sides by two, that means F two is going to be equal to half of our expression here. Now, we've solved for F two, the F two that we need to eventually get the separation between the lenses. So we can now substitute our values to get the value or to figure out the result. So when we do that, it's going to be equal to a half OK? Of one divided by the effective focal lens which was 2.083 centimeters. OK. And you know what, let's put that in millimeters since our focal lens were stated in millimeters for F one. So that would be 20.83 millimeters minus OK, minus one divided by two multiplied by 35 millimeters. OK? Raised to the power of negative one. Now, when you put that into your calculator, OK, you should get that F two equals 14.8 millimeters or we can write it as approximately 15 millimeters. So now we've solved for F two, we know F one. So now we can find the separation between the lenses cause that tells us then that there are four, the separation between the lenses L is going to be a half of the sum of that distance, which is a half of 35 millimeters plus 15 millimeters, 35 plus 15 equals 50 a half of 50 is 25. So the separation between the lenses is 25 millimeters. When we go back to our answer choices that would have been answer choice. C Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Ramsden Eyepiece

Focal Length

Chromatic Aberration

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