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QuestionsExplain briefly why the ruled lines must coincide with the centre of the table and must face the telescope. (Hint: consider the effect on the results if th...

Question

QuestionsExplain briefly why the ruled lines must coincide with the centre of the table and must face the telescope. (Hint: consider the effect on the results if this was not the case) Suppose that you are given & monochromatic light source of wavelength 600 nm; and a diffraction with 500 lines mmn How many full orders will be visible for normal incidence grating experiment? (Hint: the maximum possible angle for diffraction is 902). For the above experimental setup the angles of minimu

Questions Explain briefly why the ruled lines must coincide with the centre of the table and must face the telescope. (Hint: consider the effect on the results if this was not the case) Suppose that you are given & monochromatic light source of wavelength 600 nm; and a diffraction with 500 lines mmn How many full orders will be visible for normal incidence grating experiment? (Hint: the maximum possible angle for diffraction is 902). For the above experimental setup the angles of minimum deviation Were found for all visible ordersf orders should be visible in this experiment? (Hint: decide on the maximum angle of How many deviation) 159 in the first-order spectrum At what angle would this same color A certain color emerges at emerge in the second order if the same source and grating are used? methods should be more accurate for a 'specific order. Use & complete error Discuss which of the two analysis to back Up your answer below. Please complete these (write Annumber of exercises involving angular vernier scales are given measurement of angles next to each) for your OWn practice.



Answers

Coherent light of wavelength 501.5 nm is sent through two parallel slits in an opaque material. Each slit is 0.700$\mu \mathrm{m}$ wide. Their centers are 2.80$\mu \mathrm{m}$ apart. The light then falls on a semi cylindrical screen, with its axis at the mid line between the slits. We would like to describe the appearance of the pattern of light visible on the screen. (a) Find the direction for each two-slit interference maximum on the screen as an angle away from the bisector of the line joining the slits. (b) How many angles are there that represent two-slit interference maxima? (c) Find the direction for each single-slit interference minimum on the screen as an angle away from the bisector of the line joining the slits. (d) How many angles are there that represent single-slit interference minima? (e) How many of the angles in part (d) are identical to those in part (a)? (f) How many bright fringes are visible on the screen? (g) If the intensity of the central fringe is $I_{\max },$ what is the intensity of the last fringe visible on the screen?

In this question we have to find out the grid lines. We know that in the fraction grating set up the principle minutes. But can they find out from the following from a data scientist to he went to M. Lambda by the This is a question # one. And distance away from the grating is why equal to Dante to this situation number two and far for an angle theater which is very very very smart. Then scientists to theater approximately equal to 20 to. So the prohibition become by it went to and scientific and why become an and and divide by D. This is equation number three For two convicted plentiful maxima. The idea of maxima RM and M plus one or two consecutive maximum the other next to a DRM And the last one. So therefore the the stance is given is L into M plus one linda. I would uh the minus l M linda outta D. And for grading but grading A. And B. We can write del Y equal to and lambda. The this is a question number one and Dell by B. It went to linda. The designation number to a number of lines per centimeter we put ah and is it will do one out of the are one of the days able to end. So a prohibition began Del by equal to and uh and limiter. This is a question number three mm hmm. There will be equal to end B and linda. This is a creation number four. So dividing dividing immigration one by two. We get Dell by a divide by Del Y the equal to Hello linda. And uh divide by l linda and be So these two will get gains here and we can write and be equal to. And I want to fly by Dell. Who I be out of Dell by. Yeah. So the number of lines we get is to 1000 into Delva is given a 3.2 centimetre Divide by two pints, seven centimetres. So the number of lines, Yes, it was 2 to 2400 lines, but to me, dead.

Now I know that the resolving power is end times M, whereas to use m equals one and an is going to be 5000 lines per centimetre times 3.5 centimeters. So our answer is just going to be 5000 times 3.5. Put that in a calculator. Of course, I could probably do that in my head. Right. Anyway, the resolving power is 17,500. Um, so we want to be all right. As I look in section 36.5. I see that in order to resolve the sodium doublet you need to have a resolving power of 1000 17,500 is greater than 1000. So, yes, you can resolve that. See, we got here. Okay, So it says that we're going to use the second order, and we're resolving spectral lines that are very close to 587 point 8002 times 10 to the negative ninth power meter spectral line of ire. All right, so, um, resolving power is lambda over Delta Lambda, and it says wave links that are longer than the iron line. So that would be Ah, I'm gonna put We've links of I for wavelength of ire. Okay. What is the shortest wavelength you can distinguish from the iron line? Okay. Which I think I could really, really just change this toe. I hear. So Delta wavelength is wavelength iron over our, um, but we're using M equals two and r equals end times. M with m equals two r r is twice as much. Two time 17,500 because our is to and we used ah, one in the previous one. Okay, so let's put this in a calculator. I'm gonna call this our sub to for, uh, chemicals to Okay, Lambda I is 5 87 point I eat here zero to times 10 to the negative ninth power over our sub two, which is two times 17,500. That gives me 1.6794 times 10 to the negative 11th power. And, ah, again we wanted Let me just double check here longer wavelengths, So I need to add this to the wavelength for iron. Okay, that gives me, um, seven significant digits. 587 point eight one 70 nano meters. Okay, let's look at this was part I Look at part I I Gates, are Now all I have to do is take Lambda. Ah, I minus Delta Lynn. So we put that in there, put a negative in front. That gives me 5 87.78 34 nano meters. So the range of wavelengths Okay. And I just realized I didn't read the question right. It says that you could not distinguish so the ones that you could not distinguish go from 5 87.7834 Teoh 5 87 point 81 70 So that's the answer to II.

Hi there. So for this problem we have the following data representative in this table about the width of the sleep, which is A and the width of the central maximum in a disruption pattern. So it is produced at a distance that are gonna call our yeah from the fleet Of 1.50 m. So the question for this problem is, and the part A is if the width of the central maximum is inversely proportional to a, then the product eight times the the product between the width of the sleet and the width of the central maximum is constant independent of a. So for this data in this table we need to represent eight the graph of the product between those two with versus the width of the sleet. So and also we need to explain why is this value constant for their smaller values of A. So um this problem and deals when the light from the sleep produces an eruption pattern on the screen, we know that the first door fringe of course went um When a times the sine of detail one is equal to the Waveland and that tangent of theta, one is equal to the wave length of the first half of the wave land, of the maximum over the distance uh to from the sleep to the to the screen. So we can say that this is equal to test where r is the distance, as I said, from the sleep to the screen. Now, for part A as I explain if tita, one is a small, we can use the small angle approximation. So we can see that when the tita is so small, we can approximate that sign of data is approximately time equal to tangent dita one, which can, we can approximate that. We know it is this volume here and it is the width of the central maximum two times the distance to this. Lead to this lead from this lead to the screen. No, in that case we will have, if we put that in here, we will have that what The product between these two, It's going to be equal to um two times are the wavelength. This comes from the uh solving this when we plot that in here. So we are obtained um We sold for signs of sign of detail one. So we obtain the wavelength over a and we substitute that in here. So we obtain the wavelength over a is equal to this. So we can pass this to the other side and this to this side. And we obtain this equation in here. So that's the equation and which is constant. We know that that value is constance because we know uh the value to from the sleep to the screen, which is capital art doesn't change. We are not changing that parameter and we can we also are not changing the parameter of the wavelength. So this circumstance, that product is a constant. But if detail one is large, we cannot use the approximation that we said in here. We cannot use this for large values of the angle. And so for that part we will attain that. The project between these two parameters is not a constant. So what we are going to do is to show the garage of a times W versus A. For the data. And this problem. So what we are going to do is multiplied each of these values. So we're going to have and here this product. So we need to multiply this by by this. So you can do that. And after that you're going to obtain a set on another set of values. So let me obtain some values. So these are some of these values. You can calculate the other ones. And we need to plot this. We need to plot the product between those two versus the width of this lead in the X. Company in the access and the product in the white assets. So when we do that you can use a software to make this easier. But you can do also by hand um We are going to obtain the following. Uh huh. This I am graph right here. Um We can see that. Um. Uh huh. Yeah and it is ah for for a small shows a curb mature and that I am for a small polly for for for the for small values of A. This product is not constant. As you can see from here for small values of a. This product is not constant issuing a Suncor better, but started at this point, we can see some come stand behavior. So that's what we were talking about about this project. Now, for part B of this problem, we are asked to use our graph to calculate the wavelength of the laser light. So to do that, we see that our graph flattens out of the large values of A. So we can see that that value for some reason for and for some value of A starts to go um uh to uh to approximate to a constant value. And so that that value is approximately. I'm gonna put that value in here. But if you store by looking by values of able to see that that value is approximately. This product is going To the value one 50 sets Times 10 to the -6 meters square. Of course this means that we can apply the small angle approximation in that case. So we will see that this volume here, the one that they put in here is equal to the approximation that we did for small, for small angles. So this is equal to two times. And there are the distance to from the from the from the sleep to the screen times the wave line. So we know all of these values. So we need just simply to solve for the wavelength. So we obtain that this is the product between These two with over the uh two times the distance to from this lead to the screen, we will have um that this is 1.56 times 10 to the minus sets meters square. And this over two times and the distance to the point, Which is 1.50 m. So when we um solve this into the calculator, we obtain a body off 520 nm, which is two nm is 10 to the -9. So that's the value for the wave land that we obtain from this. Now for parsi of this problem, we are asked um what is the angular position of the first minimum in the disruption pattern from when we have a value of a. So the first bar with a has a value of 0.78 Micro m. And when the value for a is 15 0.60 Microm. Yeah. So um we can use again in this case the equation that relates in the angle with the Waveland. So the product between them with of the sleep times the angle that it made on the screen is equal to the wavelength. And we know the value for um for a. So we need to solve for the uncle. So we plot those values in here. We also know the value for um the Waveland that we obtain in part B of this problem. So we just simply need to obtain the young old one, we pass this to the other side and we take the inverse function of the sign, Which is the sign of -1. So this is the wavelength over the width of the sleet. So when we substitute all of these values we obtain, this is one sign of -1, the William, the wavelength, which is a 520 nanometers, which is 10 to the minus nine m, the width of this lead, which is um zero point 78 micrometers. So it's 0.78 times 10 to the minus six m. So when we plot this into the calculator we obtain a value of 42 degrees. So that's the solution for part Uh part um one of the part c of this problem. So for part two we are given that now the Um the width of this lead is 15. Faith then. Sorry Faith then .60 Niekro m. So Having that, we just need to simply substitute for that value. So again we're going to obtain the angle, the first angle of the first order, which is sign of -1 the Waveland, which is again the same value as before. But in here we have different with which is 15.60 times 10 to the -6 m. So we obtain that the angle in this case has a value of one .91°. So that's the angle in the case that the width of the sleet has um This value of 15.60 μm. So this is it for this problem. Thank you.

Okay, So for this question, we can use the maximum intensity you put in, which is a Santa Isaac om blunder. Unless weightless M is just the number, Santa dies the sine function and a easily with off the slit everywhere. Attention. Failure is you go to Omega divide by two over acts. Which is he with Omega Divide two X. Okay, It's a showing. The grab here. So the axe is the distance between his green and this slip. Omega is the distance, um, honest green between the two minimum on either side of central Max. Okay, So since the ankle feather in this question is really very small, that means Santa is approximately equal. Say that which is approximately equal to attention. So there who have santa is equal to omega over two X. They will have a times omega to ask. Is it, um, Lunda? So a omega over to us is, um, Lunda. I don't have a Omega is able to x kinds m blunder. That said Emily's. You want to have a Omega's, you go to to act under. And, as you can tell, a omega is equal. Some comes though sometime count Constant, which is to extender and this is the graph I drew forties question, which is the graph a omega. Where's this? A. And as you can tell, a W is not constant when the very off a is small but as the ah well over increase, you can tell that the battle of a W is approximately constant. Okay, so therefore has jumped to a question. Be well, we know it. I was going to excellent. So therefore, land I see. With eight I'll be over to X and based on graph in the table that was given from the question AWS equal to 1.56 times tend to a parting six meters work okay and then acts which is the distance between the screen and the slit is 1.5 centimeter. So therefore, what can deter me? The weapons of the laser lights which is evil? Teoh 1.56 times 10 to the power of earning six meters, where over two times 1.50 meter and it will give us 5.2. Oh, sorry. By a point. You times 10 to the power of negative meter which is by 20 nano meter is because tens of power tend to a problem. Next nine meter is he was a one on a meter. Okay? And we'll questions See, while we have two different cases. So the first case for the first angular position we have Ah, a centralized, um, blunder. So therefore, sanitizing goto m blunder. Okay, let's say Santa otherwise, um, loader, in this case, you're saying that the first men so therefore must be able to one. And we already know the weapons is fight only now meter, which is a £5.2.10 pounds meters. And we know a woman's is given zero points. I'm a Michael meter, which is some point. It has 10 departing seven Peter, so we can plan the ambassador your question to the Termini on the same day. That wall. So what does science say? That one is equal to one times by 10.2 times sent to the power of Nega seven meter over 7.8 times 10 to the power of next seven meter. And this will give us something that one is. You go to 0.667 So, uh, therefore say that one Let me see Oh, so say the one is equal to inverse sine function off 0.667 which is signed to a power of Nega Juan 0.667 and doesn't give us the angle is about for it to a degree. So this is the first angular position where half before a second case we have sat. There are two musical inland over a two. So it's talking about first minimum Salemme with one and land is the same in a To In this case is given as a 15.6 Michael meter which is one points by six times to tender part magnify meter. So they're broken determining Santana to which is equal to one times five points You times 10 to the power of Negus Abba meter over 1.56 times tens for the power off *** five meter and this will give us San say that who is equal Teoh zero point Joe three injury. Okay, so therefore or have their to is you go to in verse in function 0.33 which will have the angle weighs about one point nine degree and these are the answers for these questions


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