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A) Say what is the difference between Fraunhofer and Fresnel diffraction according to the experimental arrangement used.b) Make a diagram of the experimental arrang...

Question

A) Say what is the difference between Fraunhofer and Fresnel diffraction according to the experimental arrangement used.b) Make a diagram of the experimental arrangement that allows to observe Fraunhofer diffraction. c) Draw the Fraunhofer diffraction pattern produced by a slit of width a and of a double slit of width a and separation between them d. Clearly state the differences for each pattern.d) Calculate the interference pattern that would be obtained if the double slit in Young's expe

a) Say what is the difference between Fraunhofer and Fresnel diffraction according to the experimental arrangement used.b) Make a diagram of the experimental arrangement that allows to observe Fraunhofer diffraction. c) Draw the Fraunhofer diffraction pattern produced by a slit of width a and of a double slit of width a and separation between them d. Clearly state the differences for each pattern.d) Calculate the interference pattern that would be obtained if the double slit in Young's experiment were replaced by a triple slit. (Note that the screen is too far from the slits.)



Answers

A diffraction experiment involving two thin parallel slits yields the pattern of closely spaced bright and dark fringes shown in Fig. E36.26. Only the central portion of the pattern is shown in the figure. The bright spots are equally spaced at 1.53 $\mathrm{mm}$ center to center (except for the missing spots) on a screen 2.50 $\mathrm{m}$ from the slits. The light source was a He-Ne laser producing a wavelength of 632.8 nm. (a) How far apart are the two slits? (b) How wide is each one?

We are told that de equals three a and, um, from using the equation for diffraction of sign data equals lambda Ah, over A and I am using the equation for interference of Lambda equals X d over l or Rather, let me use sign. Data equals limbed Ah, over De and I can put an m here also. Trenary, There we go. So in part A, we're looking for the next four Maxima in the pattern, so M equals one. M equals two m equals three m equals four. Fada is the inverse sign of land over D. And so, putting that in a calculator, setting my calculator for degrees in verse sign Ah, lambda uh, over de, wait a minute. I would angles And, you know, my answer is gonna have to be in the turn in terms of Lambda and D, so I can't really put this in the calculator. It's just gonna have to be. They'd ah is the inverse sign of Landover D in verse Sign of two Lambda over D in verse Sign the re lambda over D in verse Sign four Lambda over D. Okay, so that's the answer to a Let's move one to be. So we're told to use I subzero here. What are the intensity of each of those four angles? Well, I equals I subzero. The sign of two pi over the wavelength times a science data over, um, two pi over wavelength. A sign they'd, uh, square. So the sign of the inverse sign is going to cancel out. So we're gonna have ah, I m lambda over d squared. OK, notice that the wavelengths cancel out. So that's going to give me two pi a m over D the sign of two pi a m over D over two pi a, um over de squared. All right, so that's gonna be into the intensity for M equals one, 23 and four. See which double slit interference maximum are missing. Okay, so looking at this equation, sign of data equals lambda over a and just that I can differentiate it. I am going to change this toe, fi. So if the sign of phi equals the sign of Fada, then a Maxima from here will be on a minimum from here. So the sign of Phi equals Lambda over a equals the sign of something data which is going to be Landover D No Lambda em over de so lambda over a equals Lambda em over D and then solving for m I get em equals d over a. So this is the distance between slits and this is the slit width. So if the distance between slits is an integer multiple of the slit with than those are the Maxima, that will be missing. So now I'm supposed to compare to figure 36.12 c. So let me see if I can find this 36.12. See? Okay, so I'm looking at it and I see basically this, um, this is the bottom here. Whoops. So this occurs right there and right there So the Maxima are stomped out by the diffraction curve. Um, I don't really see how my results are different other than, um, Okay, that one's his d equals four A. And this were using d equals three a. And so I m equals three. I didn't realize that until right now. So 1234 in that drawing M equals forest stamped out. But in my drawing, it just so happens that ah 0123 is stamped out. Um, whereas forest stamped out on figure 36.12 c. So should could I have been more clear. And now that I remembered that d equals three a, um, well, and see, the answer was three. Looks like I could have been more clear here. I equals ice of zero sign. Uh, if d is three a, then that's going to be 2/3. Hi. I m over 2/3 pie. And and so I could get values for all of these. So let's go ahead and do it with M equals one. The sign. And I need to put this in radiance of 2/3 times pi over 2/3 times pi gives me okay. And this also has to be squared is me 0.171 Using M is one. If I use m equals two is me 0.4 to 7. If, um, I equals three chemicals three. I mean gives me zero. If I am equal

2020 64. So we want to understand the difference between the diffraction you see using one slip versus two slits so single slipped a fraction. If this access is zero angle on our screen produces a an intensity pattern that looks like this. Now what happens when you add a second slid is you'll get the intensity pattern from one slip. But then, in addition to that, you'll have sort of more structure, uh, caused by the interference between having two sources of light now. And so it has its own little oscillation here. That sort of follows the shape of the single slit pattern. Now, as you increase the number of slits, what happens to the intensity is that it increases. And this is because you're increasing the number of sources basically of the light and accordance with Huygens principle. So the sort of number of Oregon sources is increased. Morris lets us

Belongs to the wave optics chapter. And have intensity pattern on two slit experiment. To slit experiment. Okay. As for the question. And red lines represent the actual intensity as a function of angle. And while the green line represent the envelope of the single different interference pattern. Okay, so for the part A we have to determine the slate with a in terms of lambda vigilant. Okay, so for the first minima on the either side of the central maxima, we know that a scientist to it is equal to uh a santa. It is equal to M linda and M will be one here. So we will get lambda and for the small angle is santa is nearly equal to tita. So we get 80 to that is equals two lambda. Okay, so from here we get equals two lambda divided by T. To and from the diagram we can see that data is equal to 0.1 radio. So from here we get equals to 10 lambda. Okay, so this becomes the answer for the part of the problem. Okay now moving to the part B in which we have to calculate the center to center slit separation in terms of wavelength lambda. So we can note that interference maximum for the double slit it is equal to decent it up. It is equals to um linda. And as the angle is small so we can write that the media player but he to it will be equals two M. Lambda. And from 0 to 0.1 radiant, there are 10 interference maxima which we can count so we can write that. Ah that D it will be equal to M. Lambda. They were by tita so M is equal to hear 10 RB lamda and beta is 0.1. Kita is 0.1 here. Okay so geo point so from here after solving we will get that it is equal to 200 lambda. So this becomes the answer for the part B of the problem. Okay now moving to the part C. In which using this information in the graph determine the ratio of the slit with a to the center to center distance D. We have to determine this ratio. Okay so we have obtained both the values so we get a body that is equal to tell them that they were about 100 lambda. So from here we get A by D. That is equal to one x 10. Or we can write that one ratio 10. So this is the answer for the part C. Of the problem. Okay now moving to the next part D. In which can you calculate the wavelength of the light, linda, actual separation, actual sleep separation and the slate with a. Okay, so since we do not have value for the lambda, so we cannot calculate these values. We cannot calculate these values. Okay, so these values can only be determined in the terms of can be in the term of wavelength lambda. Okay. So we cannot calculate them. Okay?

Okay. So those questions asking if we can look at the differences in detection methods for a pretty simple physical property being bond length and see if we can explain why different detection methods might give us different answers. And this is a very important question. The more detection methods and more properties you need to look into. So it uses the example B to H seven, an ion. And we're not gonna worry about Louis structure for this because it doesn't matter and it's kind of a nightmare. So let's go ahead and just look at the fact that we have a boron hydrogen bond on a terminal end. And so if we look at X ray scattering, we get a A bond length of about 103 Pekka meters. And if we look at neutron scattering, we get a length of 118 Pekka meters Pekka meters. Okay, so both of these are incredibly accurate methods of measuring bond length, Um, to the point that X ray diffraction is used as the way that we know how whole proteins look. Onda many other molecules and neutron scattering is probably more accurate, but it's very expensive. So we need to know. We can't say that one of them is just fundamentally less certain than the other. They're both very, very good, uh, methods of detection. So we need to examine why, for this example, we see a different length for this bond. So let's go ahead and look at that. If we have an Adam, here's the nucleus. Here's the core electrons, and we're gonna say there's another one here. Here's its core electrons and they're bound with their valence electrons, right? So, nuclei, I'll do this for nucleus for a neutron are going to come in and be deflected by the nuclei, right? And so, using this angle and this distance, which is not drawn proportionately. But this is going to be the same distance is this, and using this this'll angle is going to be the same. Using the difference in distance and angles and everything of different atoms, we can get the whole shape. It's very, very accurate and very, very good. And so nuclear I will bounce off of the, uh, new neutrons will bounce off of the nuclei, but if they passed through the electron field, they'll basically just go right through because they're not charged or anything. So if we do the same thing with, um X rays do this for X ray. Right? So they're going to come in and bounce off of the core electrons. Okay, Because while x rays are here, so this is nuclear notation and X rays are also zero. They're 00 gamma like that, and they have an Elektronik wave and a magnetic wave being radiation. So some of them will also go through the magnet, the electron, the electrons in the bonding area. But some of them will also be deflected, right? And so not very many will actually be deflected for this. But some of them will. Some of them will also be deflected from the nuclei. So we have these different ways that they're deflected. And remember the new the neutrons will Onley be deflected by the nuclear. That's part of what makes it so expensive. Because nuclear neutrons are much, much harder to control and simple radiation. Oscar ahead and redraw. This didn't want that to happen. Sorry. So Okay, so let's go ahead and look at why. That leads to a difference. Mhm. If we don't have core electrons like in hydrogen. Here's a Djetou. Then the Onley things that the electron that the X rays Kenbrell ounce off of are going to be the nuclei and the meat the mean electron field. Right. So that leads to this uncertainty when you don't have core electrons. And actually, the accuracy of X ray diffraction is going to be proportional to the the nuclear charge of so the nuclei, their place on the periodic table. So hydrogen having one is going to lead to the least accurate X ray diffraction. Um, possible. Really? You can't really have an X ray diffraction worse than for H two. And if we look at what boron Boron has a z of on Lee five, which is really not that much better. But it does have the one s and two s core electrons. So at least that Boren has that going for it. But you really can't do um, X ray diffraction for hydrogen very easily. So that's where this difference comes from in these two. It's just that hydrogen is incredibly difficult to utilize the x ray diffraction method for Okay, so that's really it. Thank you.


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