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A pattern of fringes is produced when monochromatic light passes through a pair of thin slits. Would such a patern be produced by three parallel thin slits? By thou...

Question

A pattern of fringes is produced when monochromatic light passes through a pair of thin slits. Would such a patern be produced by three parallel thin slits? By thousands of such shes? Give an example to support. your answers.

A pattern of fringes is produced when monochromatic light passes through a pair of thin slits. Would such a patern be produced by three parallel thin slits? By thousands of such shes? Give an example to support. your answers.



Answers

A double-slit system with individual slit widths of 0.030 $\mathrm{mm}$ and a slit separation of 0.18 $\mathrm{mm}$ is illuminated with 500 $\mathrm{nm}$ light directed perpendicular to the plane of the slits. What is the total
number of complete bright fringes appearing between the two
first-order minima of the diffraction pattern? (Do not count the
fringes that coincide with the minima of the diffraction pattern.)

Hello, everyone, This is problem 13 from Chapter 28 says consider a to slit interference pattern with ah, light of wavelength. Lambda was the path difference. Don't the l for, uh a the fourth bright fringe. And then be the third dark fringe above the central bright fringe. Give your answers in terms of the wavelength of the light. Okay, so when we have a double slit interference pattern, we have it showing up on the screen. We know that there will be a bright fringe, this point here the dark fringe, bright friend, stark fringe, etcetera, going outward. This is a central, bright fringe. It will be a dark friend and, of course, is symmetrical, like so all right. And if we're looking for the path difference, of course the path difference of the middle is zero. Because there's no difference in the distance. Traveled from the first flit in the second slit. And let's go to the second bright fringe. So this must be the difference. Must be a multiple of the wavelength lambda in order to get here, But specifically it must be one Linda, because it's the 1st 1 It's the 1st 1 out. So We know that the path difference here is actually equal to Linda. The path difference here, one more out, is dealt Jellicles to Linda, and it's the same thing on the other side. There's different lambda here to Linda here, this is Delta helical zero. Then that the dark fringes Let's talk about the dark fringes for actually enter this question dark fringes, right? Because there's destructive interference. You know, it must be 1/2 manager. And since the 1st 1 you know this one must be glammed over, too. And then someone three land over to here and then right here, be five. Land over to etcetera. So party is asking for the fourth bright fringe above the central maximum. A bump the central fridge. So see the 1st 1 above the central break Fringe has path difference. Lambda 2nd 1 to Linda, etcetera. So we know that this is just for land. Party Says wants us to give the path difference for the third dark fringe. They're dark friend joins the first dark friend just lived over to second dark fringe Freeland over too. Third dork friends, it's going to be five limbs over, too, so, you know So you can also do this with, uh So first part for maximum, you can just use em lambda. But because four there and then for the second part, you can use M plus 1/2. Linda, Um, does this actually, I suppose this does. Actually, um, it doesn't quite actually work. You have to start. M equals, um yes, sir. Chemical zero. So you can also use and minus 1/2 and then the third fringe will be m equals three. So of course, he's give the same numbers. This one M cannot be equal to zero, but they give the same things. And so if you count him out, it's quite safe. Is this a safe way to do it? So I went over to three Linda over 2 500 to bless the third dark fridge.

In this problem, we are asked to find ratio off the single slit d do there. Um, double slit separation, which is a d. We can find that by writing me, Krystle for each slipped for single slippery here. Uh, P is equal to the d. Sign that sign that Just be cool, too. End times, Linda, for a single slip. We have an easy call to one. Then our D becomes a D is tickle, too. Um, so do you're so d is equal to do you sign that is equal to lend up for the double slit we have m is equal to five. Then we could write a small D scientist up is equal to five times off, 11 a, then dividing these two numbers here we have a D sign later is equal to Linda and then divide it by small design data scientist, which is, um, five Linda. This gives us the cancel in sign sign because of the same England, Lendl and I. Same. So they would cancel out. So d, divided by small D is won over five sleep separation is a five times. Then they slid worth here

Eso in a double slit experiment, the largest data can be 90 degrees. So if you want to find the most possible number off bright fringes, we need to say that what happens if data is equal to 90 degrees and sign up? Data would be equal to one and then m lambda over D physical 21 Now we want to find em so em is equal to D over Lambda or ah, it's cool too. 0.76 times 10 to the minus six meters. Divide it by 625 times 10 to the minus nine Munis on this is a call to 0.62 or ah, number off em is equal to six.

So in this question we have a dark fringe forming at the anger. 34 degrees in a single slate Gidwitz w A and another dark fringe forming at angle 56 degrees in a slate with vets off. WP this both this dark fringes are of the same order. So ems are equal and also they are the same light. So, uh, the wavelength is are equal. This is the equation for the dark, dark fringes in a single Smith experiments. So from here, I can say sign up data A divided by sine off data B is equal to em times Lambda eight times Lambda A divided by W A divided by and B times Lambda be divided by W b. So m A and M B are equal. Lambda and Lambda B R equals So this would vehicle to let w be divided by that. Were you a eso? Since I know data a Aunt Ada be I can say w a over w b is equal to ah sign of Tater be or 56 degrees divided by ah sign off 34 degrees and and here I can find w a over w B to be equal to approximately equal to one point fine


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