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For double slit intensity pattern, formed with waves having wavelength of A = 2.2cm, the first minimum occurs at 9.5 What must the slit separation d be in the equat...

Question

For double slit intensity pattern, formed with waves having wavelength of A = 2.2cm, the first minimum occurs at 9.5 What must the slit separation d be in the equation: sine (m + 12)4"Carefully think about the appropriate value for m.

For double slit intensity pattern, formed with waves having wavelength of A = 2.2cm, the first minimum occurs at 9.5 What must the slit separation d be in the equation: sine (m + 12)4 "Carefully think about the appropriate value for m.



Answers

Light of wavelength 440 $\mathrm{nm}$ passes through a double slit, yielding a diffraction pattern whose graph of intensity $I$ versus angular position $\theta$ is shown in Fig. $36-44$ . Calculate (a) the slit width
and (b) the slit separation. (c) Verify the displayed intensitities of
the $m=1$ and $m=2$ interference fringes.

Um, I was still was still on light. We have, ah, light with Lambda is 440 Newton Media on this happens to be a double sleep diffraction. Well, given that, um, the size off the slit is a and then there's a gap between the two slits. On the size of that is D on the opposite side, we have a screen, Remember, the light moves up until the screen. And if you think about the fractions interference between the two light winds, it's gonna be an Uncle Ada, the interference uncle. And this is Ah, um, double sleet. The intensity off the lights happens whenever you have interference. And, of course, the ah leader is the Angula position. If you I want to think of it that way, that's those of the pieces of information. Well, given. The second thing that were given is a graph. Uh, these the graph on the X axis we have data in degrees on, of course, on the Y axis we have the intensity I and that's measured in terms off Millie. What's for sending me to squared? That's the intensity. The graph itself has. Ah, wavy shape multiple lives so gonna have it. The mark's mom is at seven. Merely what's percent of you two squared and then it bounces off. The access bounces off the egg sacs is you want to call it that? So it goes up and down, bounces off the x axis, comes back up again, bounces off the x axis again, comes up the third time, bounces off and then the fourth time, and then he goes on and on and on up until it's tapers off on the X axis, we can extend that, uh, the lowest point where the tempering happens is that five degrees. I want to go that go back a little bit on that. Eso This happens at five degrees, suck like a central point and then a TTE the fast point where we have the fast interference diffraction interference at them. Chemicals to one. The angle is 1.25 degrees and also the second point. It's twice that which is a M to or M equals two to where the diffraction happens. It's 2.5 degrees, so in part they wanna cultural ate the slit with. That's what we're gonna do in part. Ain't but the We wanna calculate the sleet separation, Of course. Remember, from the diagram, the slit with his a And then the separation is D uh, input. C way No wanted. Very funny. The intensities. Um, I at one. And I have to when m equals to one and chemicals to to these ones are the, uh, interference fringes. Those of the interference, huh? Fringes. So that's that's what we're looking for in the problems. I'm just going to jump on to the next page and start walking on the problems in. But they of course, they wanted us to find the slit. The slate with, um and the diffraction minimum is given us. The diffraction minimum is given us a sign Pita equals to em. Lambda eyes. The slit with lambda is the Waveland. Ah, M is the diffraction position of the diffraction interference, whether it's one or two and so on and so forth. And then, of course, the data is the angle off diffraction. Um, So since a sign data equals two m number, we can solve for a by dividing both sides by not by sign off data. So it becomes m Lunda off a sign of data, which is one 4 40 times one meter over 10 to the 9 90 meters. All of a sign off either when m equals to one. Um, sign off data if you go back. This is Yeah, this is five degrees. Okay, this is five degrees. So that happens. Um, when When m equals to one. And that's the number we're going to use in the problem. So this is gonna be sign off. Sign of five. Ah, sign of five degrees. That's what we're using. And so when we solve the problem, we get the slit with is equivalent to 5.48 micro meters, which can be rounded to five point hole. My commuters want to think of it, actually, and the second part of the problem was supposed to solve D in the second part of the problem. Uh, we're gonna stall for the so in this in this instance, in part B we're required to solve for D and D happens to be, uh, this gap right here that somebody already solved for air, which is the slit with now we want to solve for the so, um and he happens to be the slit separation are the number of spots is the racial ah d to A and he's asleep Separation. We have nine sports from the graph. We're gonna use that The relationship between the spots and D and the slit separation is given by, uh, the equation to D over a plus one equals 29 and gonna solve full D. In this case, the trucks one on both sides. We have to d over a equals 28 divide both sides by two Saudi over a course to four and then multiply able size by a So we have the equals two for a we really saw for in the previous page Ah, rounded it off to 5.0. This one becomes, uh, the equals to fall a which is the same as four times five point or micro meters. So finally, we have a solution for this problem that d becomes 20 Bikram eaters. That's this part right here. The separation in the last part of the problem, we're dealing with intensities. So, uh, this is But see, um, we're dealing with intensity off diffraction, uh, for the uncle data. And so I data equals two. Hi mugs. So enough over offa squared Where I am is the maximum intensity. Of course we have to solve for alpha and Alpha is given by, uh, a different equation. Different expression. So alpha equals two. Um, pie Lambda Science data number is the weapon. And is the sleep separate or sleep with? And then there is the angle. Ah, diffraction, uncle. So then we're gonna plug in high times five point old times, 10 to the negative six readers. That's how it and we solved that. This part of the problem micro and then sign off data. Uh, data at M equals to one is right here. 1.25 degrees. This is data chemicals to one. And that's what we're gonna use 1.25 off Lambda 440 ah, times one meter overturned to the nine. And so when we solve this, we find that the radiance of 0.7787 radiance That's so alfa. Now that we know offer gonna jump on to the next page and soulful I want If you want to call it that I am sign offa all over. Offer squared and bowling the Max is seven. Ah, Milly, What's worse? Enemy to squared. Sign of offer. We solved Offering the last page 0.7787 radiance. 7787 Radiance. We just want to make sure that those numbers are right. Um and then 0.7787 Radiance helps to have a parenthesis right there. I want to square that and soul, we end up with 5.695 Really? What centimeters squared We rounded off when an equals to one. Um, our chemicals to one. The intensity is 5.7. Merely What's presenting two squared? That's a low intensity. The next part. We want to find the intensity one m equals to two. We go back to the diagram and see their toe on chemicals to to the angle is 2.5 degrees. And that's the younger we're gonna use We need to solve for our fall For that angle. This is a sign date over Lunda by a well, really solve. For that times 10 to the negative. Six meters signed 2.5 degrees. We go back, we see that's the uncle with Dylan. Would Ah, Lambda is 440 Naina leaders changed that two meters. I mean, this is 1.557 radiance. The next page we can pledge in for the intensity it too. We know that it's I'm a sign of offer over our phone you to square that these stays at seven. Merely what sending me to squared? We square that this is sign off 1.55 seven radiance, which is what we got in the previous page. We're using it to solve this problem as Alfa always in radiance. And then we're going to square the result. We end up with 28862.886 million watts. Centimeters squared. So when M equals to two, the intensity acts chemicals to two is rounded off to 2.9. Really? What's presenting two squared? That's the intensity. So I hope you enjoy the problem. We had two parts to the problem to initial parts to the problem. The fast one is too soft for the sleep separation. Then the the slate, Witten and the slit separation. And finally, the intensities. When they interference, fringes happened to be at someone and then too used the diffraction minimum to solve for the slate, with which is five point a rounded and then to solve for slit separation. We had to come up with a new relationship that isolated D and ended up getting 20 micrometers. Then we had to solve the intensity. First of all, getting the alphas at specific Hey, does the fast data was at 1.25 are the second data. The fast data gives us 15.7 micro or rather, Millie. What's presented me to square. And then the second data is 2.5 for M two with an uncle of 1.557 radiance. We end up with an intensity of 2.9. Well, uh, Millie Watts per centimeter square. So hope you enjoy the problem. Feel free to send any questions or comments and have a wonderful day.

For this problem on the topic of refraction, we have 440 nanom light passing through a double slit and gives a diffraction pattern whose graph of intensity versus angular position is shown in the figure. We want to calculate the split with the slit separation and verify that the displayed intensities for a musical to one and a musical to match. Yeah. Now the first minimum of diffraction pattern is at five degrees. And so A. Is equal to lambda over scientific to Which is the wavelength of light. zero full for micrometers divided by The sign of 5°. And so we get the slit width to be 5.05 micrometers. Yeah. Mhm. For part B. The fort bright fringes missing. And so the is equal 24 A. And The slit separation d. is there for four Times 5.05 Micrometers. Which gives the separation between adjacent slips. To be 20.2 micrometers. For part C. For the M. Is equal to one bright fringe. We have alpha is equal to hi a sign theta divided by blender, which is pi Times 5.05 micrometers times the sine of 1.25 degrees over zero point full full micrometers. And so we get alpha two b 0.787 radiance. And so consequently the intensity of the musical. The one fringe i is equal to the maximum intensity imax into sign alpha over alpha of squared, which is mhm seven milli watts per square centimeter times The sine of 0.787 radiance Divided by 0.787 old squared, which gives the intensity Of 5.7 million watts per square centimeter. Yeah, okay. And similarly for M is equal to two Mhm. If we perform this equation again, we get the intensity I to be 2.9 milli watts per square centimeter. And both these values are in agreement with the figure

For this question was considering a double slit experiment where the central diffraction Pete contains 13 interference fringes. How many fringes are contained within each secondary diffraction peak between m equals plus one and two and equation to assuming the first diffraction minimum occurs in an interference minimum de scientist equals and lambda for em. Of course, it's plus or minus +123 and so on. An integer values. Okay, so given the central diffraction piece contains 13 interference fringes, we know that the breath of interference fringes occur. Wyn de signed data is equaled in Lambda and here, based on what we were given D is equal to 13 over to because it's half of the amount times the distance between a. Okay. So again, since there are 13 bright fringes on each one on either side of the central positions, you have divided by two. Then for the diffraction first minimum is gonna occur at an angle that we can call fate a one So we can say when m is equal to one. We have de times a sine of the angle fatal one. It's just gonna be equal to Lambda. Great, because Emma's a little one. And the 2nd 1 it asked us to find was for M is equal to two. So and then uh uh, this is a not D here says a distance, eh? Now, how about when M is equal to two When M is equal to two, we have a times The sine of the angle fated to is equal to two time Calamba because in was equal to two. So now we can calculate the values for of em, for which, uh, let's see, it will start a new page. We have conditions we have fatal. One is less than the arbitrary angle. Fada is less than they did too. Then we're gonna have. That means that the sine of the angle fatal one is less than sign. Data is less than sign data, too. Well, we have equations that have science data sine theta one and signed data to in it so we can simply solve those equations for scientist a one signed data and scientist to two and plug them in. And we find that we have Lambda over a is less san two times and claims Lambda, divided by 13 times a, is less than two Lambda over a. Well, we can go ahead and cancel the Lambda Sze in the A's out because there's a lambda over a in every single one of these. So what we end up with if we do that, this simplifies down to one is less than to over 13 times in is less than two. So now we can imagine values of M, which would make this true again. M has to be an integer. So if M is equal to, let's say, Let's just start out with one. This isn't true because two thirteen's is not greater than one in less than two. Actually, it turns out the lowest value of M that is greater than one and less than two is seven and the highest value of M. That is greater than one in less than two. If we plug it into this, expression is 12 so we have 78 nine, 10 11 and 12 as all the values of em that are acceptable that will satisfy this expression. So based on that information, we can answer how many, uh, fringes are contained within each secondary diffraction peak. You can say that there are six fringes between each secondary diffraction mission. That's high spirit. Okay, we can go ahead and box it in is our solution to the question.

Hi, everyone. This is the problem based on interference off light by double sell it in this problem, it is given light off wavelength 6 43 nanometer that is 10 to the power minus 9 m interfered through the two slits. Having the separation 20.15 millimeter. We have to find a part of the French for angle off interference, Toby, while 737 degrees So part difference is defined. Its design off theater. Yeah, the is given point tend to depart. Ministry sign up 0.737 So it is to be 1.93 and tow. Tend to the par minus 6 m. No, this is to be compared with violent off light. So data by Lambda is called toe 1.93 10 to the power minus six. Divided by babe. Length off light is given just a moment. 6. 43 and tow 10 to the power minus night. So it is three. So we can say part of France's three times off prevalence since. Okay, party friendship. Mhm and Kriegel, multiple off. We have let off light. So constructive interference take place. And Maxima Bill occurred. That's all. Thanks for watching it


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