You are given a slide with two slits cut into it and asked...

You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen $3.40 \mathrm{m}$ away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by $7.00 \mathrm{mm} .$ What is the separation of the two slits?

Transcript

00:02 All right.

00:02 Looks like you have a question about a diffraction problem where you're shining white light through a double slit diffraction grading and you're getting the rainbow effect.

00:15 And the information given was they compared where the bright fringe is beginning to happen and what width that has from the red to the violet.

00:29 They gave that at seven millimeters.

00:31 So i changed that to meters.

00:33 I'm going to change everything to meters because the distance to the view screen is in meters.

00:38 So we got to have a consistent measurement for length.

00:42 I also change the wavelengths of red and violet light to 700 times 10 to the negative 9 meters because nanometers means times 10 to the negative 9.

00:55 And then we're going to find everything from the first magnitude.

00:59 So m is going to be one.

01:00 So the equation that we're using at the top, the diffraction equation.

01:04 That the separation distance from the central fringe, that's what y stands for, is equal to m, the magnitude, which is going to be one in this scenario, times the wavelength, which we have red and violet wavelengths, times the distance to the view screen, which is given in the problem at 3 .4, divided by the slit width, which is what we're solving for, lowercase d.

01:33 Now, thankfully, they gave us that the relationship between what's happening between the two, the red and the violet, where are they located in terms of their bright french locations.

01:46 We know that the difference between their bright french locations is 7 millimeters.

01:52 That's what we're going to use to make a substitution.

01:55 Basically, we had two variable or three variables missing, the y's and the d.

02:00 So we have to make a substitution.

02:05 We do have three equations.

02:08 We can find the y for the red and the violet if we knew all of the variables.

02:13 We also know that the difference between the two ys is 7 millimeters.

02:18 So we actually have three equations and three variables, but we're going to be able to solve this by just making the substitution that i did right here.

02:29 The y value for red, the y minus the y value for red, violet must equal 0 .007, or 0 .007, i should say...

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