Prove n1<N<n2 for effective index N....

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Alvis
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Homework Statement


Prove for effective index N that n1<N<n2.

Homework Equations


[/B]
N=n1sin(theta)
TIR is theta>thetacritical
snells law-n1sin(theta)=n2sin(theta2)

The Attempt at a Solution



I know why N is strictly less than n1 since sin(theta) goes from 0 to 1 and if its at 1 theta has to be 90. For TIR to actually happen N must be strictly less than n1. But I'm having trouble proving the n2<N part.
 

Discussion

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Alvis said:
I know why N is strictly less than n1
Yet your "to be shown" has it > n1. Reversing that doesn't help since from Snell's law it should also be less than n2.
Please describe the set- up and define effective index. Even if it is a standard term, many on this forum would need to look it up.
 
Ah, my mistake. It should be show n2<N<n1.

Effective index is n1sin(theta). This is for the symmetric, 3-layer slab waveguide. The core thickness is d and its index is n1. The clad indices have the same value of n2.
My task is to prove the effective index N of any of the guided modes obeys the relationships n2<N<n1.

I know N<n1 because in(theta) goes from 0 to 1. If it is at 1, theta must be 90 degrees, meaning the light never hit the surface in front of the plane. Therefore, for TIR to actually happen, N must be strictly less than n1.

My apologies!
 
Alvis said:
Ah, my mistake. It should be show n2<N<n1.

Effective index is n1sin(theta). This is for the symmetric, 3-layer slab waveguide. The core thickness is d and its index is n1. The clad indices have the same value of n2.
My task is to prove the effective index N of any of the guided modes obeys the relationships n2<N<n1.

I know N<n1 because in(theta) goes from 0 to 1. If it is at 1, theta must be 90 degrees, meaning the light never hit the surface in front of the plane. Therefore, for TIR to actually happen, N must be strictly less than n1.

My apologies!
Ok.
Snell's Law is for a wave which penetrates the boundary. Your wave at angle theta is to be reflected.
If N<n2, what will happen?