To stay within your budget you are building a spectrometer...

To stay within your budget, you are building a spectrometer with a spectral resolution of only 100 to work near 0.9 mm. You will be using the CCD described in exercise 3.4, which you got cheap on eBay. You should have found in that exercise that the fringe period is about 0.0076 mm. If your goal is to measure emission line strengths, do you need to increase your budget to build a higher-resolution spectrometer?

Transcript

00:01 We're asked to use a diffraction grading.

00:04 So in a diffraction grading, the equation is m -lamda equals d -sign theta.

00:21 And the tangent of theta is x over l, where l is the length to the paper.

00:45 Now, we're normally given for a diffraction grading the number of lines per meter.

00:54 So d is going to be one over, i'm going to call this lowercase l, where lowercase l is lines per meter.

01:16 So now i have sine theta over lowercase l.

01:30 Putting these equations together, i would get m -lamda equals the sign of the inverse tangent of x over l over l.

01:55 So if we know how many lines per meter, we know the distance from the paper, then we could relate the distance from the center, the distance of the peaks from the center, to the wavelength.

02:19 Maybe i am shining a laser, and let's just say that my laser has a wavelength.

02:33 Of 400 nanometers, and i shine it onto substances that have their electrons excited and then emit in response wavelengths between 400 and 700 nanometers.

03:01 That would be visible light.

03:04 Well, if i want to be able to see both 400 and 700 nanometers, that would be visible light.

03:06 Well, if i want to be able to see both 400, and 700 nanometers, then, and i'm going to leave mb1, actually i want to look here, where m is 1, then for 700 nanometers, x over l.

03:38 Let me think about, let's say that i want my x over l to be, i want my x over l to be somewhere around one like one meter and two meters, then my number of lines would have to be, let's put this into a calculator, sign of inverse tangent of one half times 700 nanometers, right, divided by 700 nanometers, my bad.

04:52 So then i would have to have 640 ,000 lines per centimeter.

05:22 That seems too high.

05:25 So i'm just going to change this, not one meter, but let's say five centimeters.

05:42 So if i change that to 0 .05.

05:50 That would now give me 71 ,000 lines per meter.

06:06 Because yeah, i'd be in lines per meter times, let's not do that.

06:21 So that's going to give me per centimeter.

06:24 It would be 7 ,000, no, it would just be 710.

06:36 So if there's 710 lines per centimeter, that's going to give me 71.

06:43 1 ,000 lines per meter.

06:56 Okay.

06:57 And let's see what happens with 400 then.