- #1
GreatEscapist
- 178
- 0
Inferometer problem: URGENT
After trying to do this problem all night the night before it was due, and no avail on just scouring the internet, I would really appreciate some quick help if anyone is still up. (Or early in the morning)
Optical computers require microscopic optical switches to turn signals on and off. One device for doing so, which can be implemented in an integrated circuit, is the Mach-Zender interferometer seen in the figure(Figure 1) on the next page. Light from an on-chip infrared laser ( wavelength = 1.080 nm) is split into two waves that travel equal distances around the arms of the interferometer. One arm passes through an electro-optic crystal, a transparent material that can change its index of refraction in response to an applied voltage. Suppose both arms are exactly the same length and the crystal's index of refraction with no applied voltage is also 1.564.
What is the first index of refraction of the electro-optic crystal larger than 1.564 that changes the optical switch to the state opposite the state you found in part a?
I don't know- my teacher barely covered this, and not very well. This problem is not like any of the others in the book.
(2*pi*Δx)/λ = 2*pi*m for constructive, where Δx is the path length difference
(2*pi*Δx)/λ = 2*pi*(m+.5) for destructive, where Δx is the path length difference
λ/n = λ_n
I know that it comes out dimly (from a previous problem). I want to then make this constructive, however, I am completely and utterly stumped at how to do that. A hint given to my class by my professor on the school forum was:
I started to set up a similar problem today in class - what you want to do is look at the path length difference - including the fact that some of the path is inside the material so you have to shift the wavelength to lambda/n (where n is the index of refraction). You're told that the paths are the same distance, so the only difference is the length it goes in the plastic - so you're comparing light that doesn't go through plastic to light that goes through plastic: x/lambda compared to x/(lambda/n) where x is the distance traveled that is different for the two rays (one in plastic and one not in plastic)
So we know that it only goes through the plastic...but how do I compare? I'm so lost...
After trying to do this problem all night the night before it was due, and no avail on just scouring the internet, I would really appreciate some quick help if anyone is still up. (Or early in the morning)
Homework Statement
Optical computers require microscopic optical switches to turn signals on and off. One device for doing so, which can be implemented in an integrated circuit, is the Mach-Zender interferometer seen in the figure(Figure 1) on the next page. Light from an on-chip infrared laser ( wavelength = 1.080 nm) is split into two waves that travel equal distances around the arms of the interferometer. One arm passes through an electro-optic crystal, a transparent material that can change its index of refraction in response to an applied voltage. Suppose both arms are exactly the same length and the crystal's index of refraction with no applied voltage is also 1.564.
What is the first index of refraction of the electro-optic crystal larger than 1.564 that changes the optical switch to the state opposite the state you found in part a?
Homework Equations
I don't know- my teacher barely covered this, and not very well. This problem is not like any of the others in the book.
(2*pi*Δx)/λ = 2*pi*m for constructive, where Δx is the path length difference
(2*pi*Δx)/λ = 2*pi*(m+.5) for destructive, where Δx is the path length difference
λ/n = λ_n
The Attempt at a Solution
I know that it comes out dimly (from a previous problem). I want to then make this constructive, however, I am completely and utterly stumped at how to do that. A hint given to my class by my professor on the school forum was:
I started to set up a similar problem today in class - what you want to do is look at the path length difference - including the fact that some of the path is inside the material so you have to shift the wavelength to lambda/n (where n is the index of refraction). You're told that the paths are the same distance, so the only difference is the length it goes in the plastic - so you're comparing light that doesn't go through plastic to light that goes through plastic: x/lambda compared to x/(lambda/n) where x is the distance traveled that is different for the two rays (one in plastic and one not in plastic)
So we know that it only goes through the plastic...but how do I compare? I'm so lost...