In a Young's interference experiment, the two slits are...

In a Young's interference experiment, the two slits are separated by 0.170 mm and the incident light includes two wavelengths: ?1 = 540 nm (green) and ?2 = 450 nm (blue). The overlapping interference patterns are observed on a screen 1.41 m from the slits. (a) Find a relationship between the orders m1 and m2 that determines where a bright fringe of the green light coincides with a bright fringe of the blue light. (The order m1 is associated with ?1, and m2 is associated with ?2.) m2 / m1 = (b) Find the minimum values of m1 and m2 such that the overlapping of the bright fringes will occur. m1 = m2 = Find the position of the overlap on the screen. cm from the central maximum

Transcript

00:04 All right, looks like you have a question about a pretty challenging diffraction question.

00:11 Up at the top left, i put the diffraction equation that we'll be using throughout.

00:16 It's the y is represented by what bright friends we're looking at, whether it's the first, second, third, and so on, order.

00:26 M is the order that we're looking at.

00:29 1, 2, 3, 4.

00:30 It has to be a positive integer.

00:34 Lambda is the wavelength of light that we're looking at.

00:39 Capital d is the distance from the view screen to the grading.

00:43 And then lower case d is the slit width.

00:45 So this is the equation that we're going to use.

00:48 And they gave us a variety of components here that we're going to be able to use.

00:53 So off to the right, i put, well, let's see, what do we have to do? we have, basically what we're going to do is we're going to make these ys equal to each other in the end.

01:04 But first, i rearrange everything to figure out what's the ratio need to be.

01:10 So if i use the diffraction equation and i made my m1, my ones be the green laser, and then the twos are the blue laser, i rearranged everything to solve for what, what's the m2 value or what order do i need for blue divided by what order i need for green.

01:34 So that's where the m2 divided by m1 is coming from.

01:38 So over here on the right in green, i saw for m1 and m2 just rearranged the variables.

01:44 And then i plug that into the ratio down below of m2 over m1.

01:50 So i ended up with this really complex fraction.

01:52 In red, i canceled out the common terms that are on both the numerator and the denominator.

01:58 So all the ds go away.

02:00 And then i rearranged it using my, when i divide by a fraction, i can flip -flop the denominator.

02:08 So i ended up with the bright fringe of the blue divided by the wavelength of the green times wavelength of the, excuse me, i mean, the twos are the blue...

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