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Anderson Video - Single-Slit Diffraction Example

Professor Anderson
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 Okay, let's continue. So, number six says a beam of monochromatic light passes through a slit that is 11 microns wide, if the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31 degrees away from the center line, what is the wavelength of light? Okay, so this is back to a single slit problem. And the idea is this, we have a single slit, some plane wave coming in, creates this diffraction pattern, and we know that- we know the slit width. I think we were calling it- what were we calling it, d? I think I called it w in in lecture but that's fine, we can use d. And then they tell us that d is 11 microns, which is 11 times 10 to the minus 6 meters. And they tell us the angle to the first dark fringe. So here's the center line, and they tell us this angle right here. And that theta is equal to 4.31 degrees. Alright, they are looking for lambda. So, how do we do this? Somebody give me some thoughts. Jessica, what should we do? Okay, d sine theta equals m times lambda. So where did that come from? Okay. Okay, but this equation that you just told me, that was for the bright fringe from a double slit. Right, and now we have the dark fringe from a single slit. So, is this equation the same or is it something else? Anybody have a thought? How do you deal with this? Let's say you're in the exam tomorrow. What are you going to do? You're going to use the same equation and just hope? No you're not. You're going to go to the formula sheet that I posted online for you guys, and you're going to look at that formula sheet and see if that equation is right or not. So let's pull up the formula sheet off of Blackboard and take a look at what it says. Okay so, you scroll down a little bit on the formula sheet, and we see the bright fringes from Young's double slit certainly obeys that equation d sine theta equals m lambda. And if you scroll down a little bit more, hopefully, won't let me scroll down. There we go. We get to single slit diffraction, which says the dark fringe is a sine theta equals p lambda. That is, in fact, the same equation as this. Okay, in that equation a is the width and so that's d, and p is an integer, that's our m. So this is kind of weird, right? But this is the bright fringe for a double slit, it's also the dark fringe for a single slit. Alright, good. So if we have that equation then we might have everything we need, right? Because Jessica said 'well, let's just solve this for lambda, that's what we're looking for.' Lambda equals d sine theta divided by m. We have d, d is 11 microns. So that's 11 times 10 to the minus 6. We have the angle, 4.13 degrees. So we can take the sine of 4.13- or, 4.31 degrees. And then we have m for this case, because in the problem they say it's the first dark fringe. So m is, in fact, equal to 1 here. Alright, punch in all these numbers in your calculator and tell me what you get. 8.26 times 10 to the negative 7 and the units are, of course, meters. And if we look at the answers all the answers say nanometers, so we need to convert this to nanometers. And if I move this over 2, then I have to subtract 2 from the exponent, and 10 to the minus 9 is a nanometer. So was there something after the 6? So this is 826 nanometers. And if you look at the answers, the closest one is answer A, 827 nanometers. Okay, so there's probably something else.