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.