3. For an asymmetric Mach-Zehnder Interferometer (MZI), a....

3. For an asymmetric Mach-Zehnder Interferometer (MZI), a. Write an expression for the square magnitude response for port 1->2 assuming 48/52 power splitting ratios for the couplers. b. If the FSR=50GHz, what is the period in wavelength? What is the relative path length difference in mm, assuming $n_e$=3.0?

Transcript

00:01 Question 13 wants us to calculate the phase difference between wavefronts under different conditions.

00:07 This is going to require an illustration from figure 37 .5 in the text.

00:14 Let me pull up my illustration of that figure in the book here.

00:19 That will help us in solving the problem.

00:22 This is a four -part question.

00:25 Part a asks us to calculate the phase difference when the end - angle theta is equal to 0 .5 degrees.

00:34 There are a few constants known to us that will be helpful for solving all parts of this problem.

00:40 L, the distance from the slit screen to the projection screen, is known to be equal to 1 .2 meters.

00:48 The slit spacing, d, is equal to 0 .12 millimeters.

00:56 And the wavelength of the light, lambda, is equal to 5.

01:00 900 nanometers.

01:02 Now the equation for phase difference denoted by the greek letter phi is equal to 2 pi over lambda times d times the sign of our angle theta and it's measured in radiance.

01:18 So in answer to the first part of our problem the phase difference phi will be equal to 2 pi divided by 500 times 10 to the minus 9 meters.

01:33 10 to the minus 9 meters is how much 1 nanometer is equal to times 0 .12 times 10 to the minus 3 meters, 1 millimeter is equal to 10 to the minus 3 meters, times the sign of 0 .5 degrees.

01:54 Calculating that out, we find that phi has a value of 13 so that's the phase difference between the two wave patterns emitted through this double -sliped screen.

02:12 Now in part b, it wants us to determine the same value again, wants us to find the phase difference.

02:20 But now what we're given is that the value for y is equal to 5 millimeters.

02:26 The value for y is the distance between the distance between the value for y is the distance between the point p in our illustration and the central max denoted by o.

02:38 I said that our phase difference was equal to 2 pi over lambda times d sine theta, as you can see in our previous part of the problem above.

02:51 How does y come into play there? y is part of the expression for what the sign of the angle theta is.

02:58 The sign of theta can be said to be equal to y over l, or l is the distance between the projection screen and the slit screen...

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