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Light with wavelength of 560 nm illuminates double slit with the distance between the slits of 95.0 um A screen is 2.20 m behind the slitsParAWhat is the angle to t...

Question

Light with wavelength of 560 nm illuminates double slit with the distance between the slits of 95.0 um A screen is 2.20 m behind the slitsParAWhat is the angle to the secpmd-order bright tringe?AzdSubmitRequest AnswerParBHow far is the second-order bright fringe from the central maximum?AdE)SubmitRequest Answer

Light with wavelength of 560 nm illuminates double slit with the distance between the slits of 95.0 um A screen is 2.20 m behind the slits ParA What is the angle to the secpmd-order bright tringe? Azd Submit Request Answer ParB How far is the second-order bright fringe from the central maximum? Ad E) Submit Request Answer



Answers

Two narrow slits $50 \mu \mathrm{m}$ apart are illuminated with light of wavelength $500 \mathrm{nm} .$ The light shines on a screen $1.2 \mathrm{m}$ distant. What is the angle of the $m=2$ bright fringe? How far is this fringe from the center of the pattern?

In this problem, the small angle approximation is valid. It's valid because this ratio here is small. It's very small, in fact. And when this ratio is very small, then we can use why ohm is equal to our and when, UH over a which is only valid sometimes in cases where this is true and now they want the distance between the two dark fringes on either side of central maximum. And since the distance between the central maximum and one of them is, why one the distance we really want us to white one. So that's that's what we want here. Well, we could just sulfur. Why one and then double it from this formula. We know that why one is equal to our land over a Just playing into him was a good one, since we want that first ah, destructive interference and now we can plug in these values which were given in the problem. R is 3.5 meters lambda 6 33 times 10 to the minus nine meters and a is 0.750 times 10 to the minus 30 meters and so we get out 2.95 times 10 to the minus 30 meters. And then if we double that after converting this two millimeters, right, this is also 2.95 millimeters. So if we double this, we get to my one, which is our desire and, um is five point 9,000,000 years, and that's the answer.

Okey dokey. We have another single flit reflection problem. The wavelength of the light that's going through our split 600 nanometers Nano meters. The screen is a distance three meters away. And we know that the wits between the first dark fringes is 4.5 millimeters. I'm gonna call that too. Why? One for a reason. Im about his plan. Reno. All right, so this is our screen. The diffraction pattern that we see on the other side is gonna look a little bit like the swollen black. So am I. A big, fat, bright fringe in the middle. I have my dark fringes right here. Okay. The distance that they give us for this problem is the distance between the two dark fringes. Why? One is the distance from the center of the bright peak to the dark. Cringe. So that's why I put two times why one? Because it's just twice the distance company in a suite. All right, so our slit and our fringe distance is small compared to the distance of my screen, so I can use the small angle approximation that tells me that the why EMS are equal to our M lambda over. Hey, where I am is the distance between the center of the break fringe to your respective dark fringe is So This is why I want This is why two by three so on, right? And we want to know what is the width of this like. So if we solve this for a is equal to our and Lambda over why? And so in this case, we have m equals one one one. So I had a street plug this in our his three meters lambda was 600 Nana meters. So 600 times 10 to the minus nine meters And why one is half of that? So pull 4.5 over too. Millie. So 10 to the minus three meters. Right? So if I plug that into my Hindu Danny calculator, I get it Distance, eh? Is equal to 0.8 millimeters, you know? Yeah,

What else? To give a problem. The angle from the central maximum to go first. Fast, bright, bright friend is bridges is given by D Sign their top, right, Right. Physical tow n times a Linda from here. Ah, fate up for first bright fringes and musical toe One for first Brian Fringes right Will be signing worse. Oh, for M times Linda divide by D Some student your values we have here Sinan Worst musical one. So Lamda easy trees 6 33 uh derided by 1.45 times Don't bother. Minus minus five. This gives us an angle for the first right fringes to be, um, heard off 2.5 degrees. But be the angle from central max to the second right second darks only know the bright dark fringes is given by again the formula that he signed Do you sign data? Dark is equal to M plus one or two times. Linda, where I am is a call to one for the first our trainees tools of student value in Seoul for a better so they had the dog will be signing worse off one plus one or two. So shooting the value for Linda, that is 6 33 times 10 to power minus night, derided by 1.45 times 10 to power minus five. So trying this time we get the data. Ah, the angle from the central max to a second doc. Fringes to be 3.75 degrees. What? See, uh, the distance off. The first bright fringes from the central max is given by why bright is equal to lend at times. L divided by tee times him where I am is a call to one we confined. Why? So why will be why bright will be Lambda, which is a 36 33 times than our minus nine times two, divided by 1.45 times didn't about minus five times in musical. The one this uses Why to be 8.73 centimeter. The distance from the first bride fringes from the central max is 8.78 point 73 centimeter

Hello. So this is a single slit experiment. And okay everybody's expression lander which is the equivalent equals X. Over D. A. Is the weight of the sled. It's is the distance between the central friends. My first dark this one picks and D. Is the distance between the slit and the screen which in this case is three m. So you're trying to find A. So if you isolate A. It will be uh you could rely under the over X. So we have longed up which is 609. Oh we just lost our extensive negative. I'm have T. B. Two B. Three. And then X. So you see the X. Uh So X. Is the distance between the red and the black right? But were given the distance between two blacks. We just have to divide by two. And that will give us uh 2.25 and that's 2.25 millimeters. Soon as I understand it the negative three. So we just put all this in the calculator. Let's see what this gives us. Uh huh. So you have 600 That's what I live it. Every line times are divided by 2.25 It's been a little train. So the width of the slit, which is a this eight. Uh huh. Times tend to negative four. I mean it's so that's the word. I would have slipped. Thank you very much.


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