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Q.2 a) A Luneburg lens is a gradient index sphere of radius R, see figure below, with the refractive index n(r) = [2-(r/R)]I2 where r is the radial distance from th...

Question

Q.2 a) A Luneburg lens is a gradient index sphere of radius R, see figure below, with the refractive index n(r) = [2-(r/R)]I2 where r is the radial distance from the centre 0. This lens focuses all rays from an infinitely distant object to a single point F'_Find the optical path length [FOF' ] inside the lens. [2 marks] Find the numerical aperture and f-number at the maximum beam diameter of 2R. [2 marks] (iii) What kind of aberrations cannot be corrected in this lens? [2 marks]b) Sket

Q.2 a) A Luneburg lens is a gradient index sphere of radius R, see figure below, with the refractive index n(r) = [2-(r/R)]I2 where r is the radial distance from the centre 0. This lens focuses all rays from an infinitely distant object to a single point F'_ Find the optical path length [FOF' ] inside the lens. [2 marks] Find the numerical aperture and f-number at the maximum beam diameter of 2R. [2 marks] (iii) What kind of aberrations cannot be corrected in this lens? [2 marks] b) Sketch the wavefront aberration W(P, 0) for the following two aberrations, paying special attention t0 their variation along the p and $ axes: Primary coma. Primary astigmatism_ [2 marks] [2 marks]



Answers

A bi-concave lens (refractive index $\frac{3}{2}$ ) has radius of curvature $R$ for each surface. If it is immersed in a medium of refractive index 2 , it will behave as (a) a converging lens of focal length $2 \mathrm{R}$ (b) a converging lens of focal length $\frac{2 \mathrm{R}}{3}$ (c) a diverging lens of focal length $2 \mathrm{R}$ (d) a diverging lens of focal length $\frac{\mathrm{R}}{3}$

Hi. In the given problem Equation 23 0.27 off. The chapter sees that the inverse off focal land off a convex lens is and minus one bracket one by our one minus one by our two. They're these organ and art are the radio off creature off both the surfaces off this convex lands and end is the absolute reflective in bags off the glass off lens. So it is said that this lens is dipped into water. We're and one is the reflective index, absolutely protected off water, which is surrounding this lands. And and two is the reflective index off glass off which this lens is made up off. Asked for that given question. So in first part of the problem, we have tow obtain, uh, changed expression for focal length means lens makers from law after making these required changes. So, in case the lens is dipped into the water, the app, the reflecting in depths off glass lands with respect, the water comes out Toby and off to with respect to one which can be given us and to bye and one so the lens makers from law becomes one by F is equal toe and off to with respect to one minus one one by our one minus one. By are to or we can say this is and to buy anyone minus one one by our one minus one by our two. So this is the expression for the focal length for the inverse off focal length off the convicts lands. Then it is dipped in the water. And this is the answer for the first part off the problem. Now, in the second part of the problem, we have to find the focal lengths off the same convex lens. In two cases, in the first case, the lens has been put in air. So the refractive index is it's absolute refractive index which is 1.5 only, and the radio off curvature off this lens are given us 40 centimeter reach. So if urban is plus 40 centimeter, then as for sign convention are you will be minus 40 centimeters as we know this is the convert flans Here, This is our one the videos off creator off which left face. So this will be positive as it is being measured in the direction off, incidentally, and this is our two, which will be taken as negative as it will be measured in a direction opposite to the direction off. Incidentally, hence now, using lens makers, Formula One by f is going toe end minus one one by our one minus one by our two and plugging in all the non values. This is 1.5 minus one one by our one means 40 minus one by minus 40. So this is 0.5 remaining outside the bracket. Then this is one by 40 plus one by 40. Here become 0.5 into two by 40 or simply won by 40. So but focal length off this convex lens in air comes out Toby plus 40 centimeter, and this is one off the answer off. Second part of the problem now, in the second part of the problem in the same second part of the problem, the second cases when the lenses dipped in water. So now in water, the absolute reflecting lacks off glass lenses 1.5, and that off water is 1.33 So now the focal length off this convex lens will be given us one by a dash is equal to and to buy n one minus one one by our one minus one by part two So plugging in all the known values again, this is end two means 1.5, divided by 1.33 minus one one by 40 minus one by minus 40. So here it is 1.1 to 8, minus one into two by 40. So it really needs as 0.1 toe eight divided by 20 as this is cancer, these forties canceled by two. So finally, dash is the focal length of this convex lens in water comes out to be 156 0.25 centimeters. So this is the last uncertain for the second part off a given problem. Hence, it is also clear that the focal length off the lens is increased. Then it is depicted in the water. Or we can say the power off this convex lens will be reduced when it is dipped in the water. Thank you

Finland has a convex surface of radius 20 cm and a concave surface of right radius 40 cm. And it's made of glass of refractive index 1.54 compute the focal length of the lens and state whether it is a converging or diverging lens. Okay, So they are, One has a radius of 20 cm and the other has a radius of 40 cm. So both of them have, their curvature occurs to the right. So we can use this in lens equation or the lens maker equation you have won over F is equal to N -1, one over R one -1 over R two. So we're told, and is 1.54. R. one is 20 cm. R two is 40 centimeters. So solving this out for the focal length And you get a focal length of 74 cm. Since it's positive, it's converging.

A double Convex Finland has a radio of 20 cm. The index of refraction of glasses. 1.5 compute the focal length of this length in air and when it is immersed in carbon di sulfide, Within the index of refraction of 1.63. Okay, so we start with the equation, one over F. Is equal to And one over into -1. One over R. One minus one over R. Two. Okay, it's a double convex lens which tells us are one is positive, so positive, 20 cm. But our two is negative negative 20 cm. And so report E. and one is 1.5 into is one. So plugging in values and this gives us the focal length of 20 cm for part B and one is 1.5 And into is 1.63. So this gives us a focal length of negative 125 centimetres.

That they have a problem any more. We're told that we have a then meniscus lens converging men's and the radius of a bird for a radio of cultures, too heard twelve centimetres in twenty eight centimeters and the refraction of the lab report six in part A. We're told that an object is placed to left of the leads forty seven meters away, and it has a height of five millimeters. And me, you're asked to determine the position and size of the image and then, in part B, a second converging lens with the same focal like there's the first times is placed at a distance, which I called the A three point one five meters, or three hundred fifty centimeters to the right of the first dance. Again, we're asked to determine the position in size of, in this case, the final image, and whether or not into determined the orientation of the final image, I'd say we were to repeat what was done in part be accept now that the distance between the lenses is forty five centimeters is that of three hundred fifty seven years. So let's go ahead and do part eighty. So the golden part is to find s and to find, uh, why primed. So let's first find us using equation thirty four point one eight. So we have whatever asked us one of our privacy too, and minus one one of our one. I mean, it's whatever art too, And meniscus lens looks like Like that. Oh, so this is this Give me our Juan and could be our too. So we know, and we know s we are one we know are too. So all that's left to do is find, find, um, best crime. So this is actually one over half thiss whole thing. So what? We just work out this left hand's hide first so we could put it in a more familiar form. Right? Then we can write whatever ass plus farmer asked. Primacy, Woods. What over half been from here? We can easily work out. That s prime is s f o ver asked me. So let's go ahead and do it. His and minus one another are one minus. Whatever art, too. And that's one point six minus one one over twelve centimeters. Minds from average twenty eight centimeters. And that should give me that should give me thirty five centimeters. So? So That that gives me that half is thirty five centimeters. You'd have to because you'LL get one over f is equal to, um here you're a point zero two eight six centimetres inverse centimeters And that new invert this thing here, you take, uh, this. Let's do whatever f able to near point there. Two, eight, six. So then half is one over zero point here, two, eight, six with centimeters on top. And that's thirty five. Senate leaders like so. And now that we have F week in years, this equation and then we could write as prime. So s I was forty. What? He five centimeters f is thirty five centimeters. We just found that. And then we have forty five. It was not his thirty five centimeters, and that gives and they give us one five seven point five centimeters. Is that now we have the position of of the image that's formed by that lens. Now we can find uh huh. So now we need to determine why primes, remember, we found asked, crime is one one five seven point five centimeters so we can use, um oh, him. Is he going to mind? Has asked prime over ass in this case in that minus one five seven points, five centimeters, divided by forty five centimeters. And that gives us Highness three point five. So and Mrs Aston zero, the images and burned it. And now we could use Amazon's that you call Tio. Why? Prime or wife? Which tells us that my prime is him times. Why? So we have three point five times the height of the object, which is five millimetres. And that gives me seventeen point five telomere. So that now is the height of the object. You know, we can a move on, too. Um, whatever. We're asked you in in part B. Okay, so now in part B, we have hit a another lands another converging lands that's placed three hundred and fifteen centimeters away from the first, So yeah, like a picture. Let's drop picture. You have your objectivity here in them. They're your first lens. And then there is your second lens. So this thing was forty five centimeters. That's our distance. That's one. And it had image, you know, um, one hundred and fifty eight centimetres away somewhere. So No, let me do that. Not a color. So it has an image of her Hair's on fire s one primed That we know is a one hundred eighty eight centimeters. You know, we have a second lands with the distance between these two lens, it's called D. That's three hundred and fifteen centimeters. Okay, so now this is asked to This is the S Do is the object distance for the second lens Clearly s do is D minus s one prime. Okay, so SDU is the separation between the lances minus the image distance the image of the first leads So that gives us the object distance of the second chance. So we need this to find as to prime so we can use the equation we're going before because we know that the second has also have the same focal ng. So the question we used before was this one and they could call these things here as one. Just so we make sure we know that we keep we keep things there that will be asked to to which is at one which is Let's just keep it is f over s to minus and now s two. That's three fifteen centimeters, minus one five, seven point five centimetres and this turns out to be one five, seven point five. It's not a meters itself so as to primed to go to one five, seven point five years times thirty five cent amuse divided by one five seven point five. That means minus thirty five Senate leaders. And we get forty five seven years. Okay, so that's the location of of the image. Now it's forty five seven years. So somewhere over here, Okay? And we need to determine the height of the object now. So that's we're running out of space. So let's do this on a new page. So first we used Hammas Eagle to minus s, too. I'm over as to as to prime. It's forty five centimeters. Yes, to was one five seven point five. That can mean negative zero point two eight six. And now, huh? Let me called us. I have to. So now I have Why Double prime. Okay, So em, too is Eagle Tio by double primer over y prime. Where? Why, Double prime? Why? Prima's what we found before. That's the height of the image from the first lens. You know why Double prime is going to be a height of the image from the second hands. So then why don't we primed is equal to him too. Terns y primed, which is zero point two eight six times seventeen points, five millimetres. And that gives me five millimeters. So remember that the object, the object height was five million years and the final image height is also five millimeters, so they have the same height. Ah, Now we need to determine the orientation of the final image. So if but so then a total magnification will be m one times M too. And this is what we found before em. One what you find that? Okay, so that's minus three point five and two is minus Jared point two eight six in this turned out to be a plus one. So since it's positive, it's erect. So the object in the image have the same orientation in the fact the magnification is one. So they the size of the image remains the same. And now we have to repeat off the stuff per part, See, Except now, Dean his forty five centimeters. So we will need s to remember asked you is to find his d minus s one primed. So the only thing that's changed now is D Because that's one Prime is still asked one primes that we have forty five centimetres minus one five, seven point five that meters and that is negative. Is he, um, negative one one, two point five centimeters. So since now this object distance is negative, we have a virtual object, so that's something to keep in mind. So we want to find as to primed again. We use the question from before, So that's as to over as to minus F. That's minus one one, two point five seven years in F is thirty five. This one Fine. That gives me D six point six nine centimeters. And as before we want so way need to find the height. So we use him. Has two prime over s to him that twenty six point six nine seven years divided by one one two point price of readers zero point zero. Say your point two three seven ham is great and zero. So this image is your act here. Well, this part, actually, um, let's say that for the end because we need Teo. Determined the orientation of the final image him. And so we have the image height. But our prime is him too. Bye, Prime. Which is your point two, three, seven times, Uh, seventeen point five millimeters. That's four point one five millimetres. And we did a term in the tunnel magnification for first to find the orientation. So remember you had a minus three. Put five for someone. And now empty. We have positive. Zero point two three seven. Uh, this is a negative number. This is last Nero, right? So the final images inverted.


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