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Construct Your Own ProblemConsider a spectrometer based on a diffraction grating. Construct a problem in which you calculate the distance between two wavelengths of electromagnetic radiation in your spectrometer. Among the things to be considered are the wavelengths you wish to be able to distinguish, the number of lines per meter on the diffraction grating, and the distance from the grating to the screen or detector. Discuss thepracticality of the device in terms of being able to discern between wavelengths of interest.

1.5 $\mathrm{m}$

Physics 103

Chapter 27

Wave Optics

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Sheffield

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were asked to use a diffraction grating. So, in a diffraction grating, the equation is m lambda equals D sign they'd And the tangent of Fada is X over. L um, where l is the length to the paper now were normally given for a diffraction grating the number of lines per meter. So d is going to be won over. I'm gonna call this lower case l Where lower case l is lines per meter. So now I have sign Fada over lower case l putting these equations together, I would get i m lambda, uh, equals the sign of the inverse tangent. Uh, X over l over. Yeah. So if we know how many lines per meter, we know the distance from the paper. Ah, then we could relate the distance from the center. This is the peaks from the center to the wavelength. Maybe I am shining a laser. And let's just say that my laser has a wavelength of 490 meters and I shine it on two substances that have their electrons excited. And then he emits in response, um, wavelengths between 400 and 700 non amir's. That would be visible light Well, if I want to be able to see both 407 100 nano meters, then I'm gonna leave em be one. I actually want to look here where m is one, then for 790 meters. Um X over l let me think about let's say that I want my X over l to be somewhere around 1/2 lake, one meter and two meters. Then my number of lines would have to be. Let's put this in tow calculator sign of inverse tangent of 1/2 times 700 nano meters. All right. Divided by 790 meters. My bad. So then I would have to have 600 40,000 lines per centimeter. That seems too high. So I'm just going Teoh, change this Not one meter, but let's say five centimeters. So if I change that Ah 05 that would now give me 71,000 lines per meter because you would be in lines per meter times. Ah, let's not do that. So that's going to give me per centimetre. It would be 7001. No, just be 710. So if there's 710 lines per centimetre. That's going to give me 71 1000 lines per meter. Okay. And let's see what happens with 400 then. Um so I want to go back to my original equation here limbed Ah. Okay. I already have it written down there. I want to figure out what X is. So taxes l tangent, beta fada. But they'd, uh, is the inverse sign of Lambda L. Okay, so put that into a calculator. Oh, which I put it to Tangent of inverse sign of lambda. Uh, no. I want to use 400 times l l waas 71,000. That gives me 5.6 eight centimeters. And what I had up here Waas X equals 5.0 centimeters. Okay, so what I have here is ah, linked to this length to the screen of two meters and ah, number of lines per centimeter of 710. And that would allow me to differentiate somewhat between 400 and 700 nanometers. Now, I think I could do better by, um, making l larger lower case l larger. Um, maybe I could double that. So let's see if I doubled it. And Ellis still too. Then I want to see what happens here. Two tangent, Inverse sign of Lambda. Uh, times L. I would give me 11 11.4 centimeters at 400 nanometers, and that's with the wavelength of 790 meters. It would be 19 point 97 There'll be 20 0.0 centimeters. So I like this more so instead of Oh, wait, that's that's supposed to be down here. Yeah. So instead of using the 710 lines per centimetre, seems like I get a lot better resolution if I use 1420 I intended to use 20 ah lions per centimeter at two centimeters away. That's going to give me differences. Ah, on the paper of 11 7 centimeters to 20 centimeters. And so this would now be more practical because, um, it's more measurable than with the 710 limes lines

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