1. PHYS 408 Applied Optics (Lecture 8)
Jan-April 2017 Edition
Jeff Young
AMPEL Rm 113

2. Quick review of key points from last lecture
One can fairly easily keep track of all multiply reflected wavefronts in a thin-film transmission problem involving incident plane wave, including both their amplitude and phase.
Summing the infinite series of interfering waves yields an interesting, non-trivial wavelength dependence to the transmission and reflection properties of the film.
The most natural quantities to describe the behaviour include the Fresnel reflection and transmission coefficients associated with both interfaces, and the thickness of the film.
Thickness of the film is crucial due to phase accumulation as the waves propagate in the film.
Ask why the thickness is important: Phase accumulation.

3. A reminder Infinite series (homework, very intuitive)
More powerful approach
Remind them of what they did last week at this stage for the single interface problem
Write down expressions for the total E field in each regi, use Boundary Conditions, solve.

4. The solution
Show pdf of handwritten algebra

5. The solution Condition (equation) for zero reflection?
d=600 nm
d=600 nm
Are these r and t from notes, previoius equation? No, they are |r|^2 and |t|^2 What do you note about these results?
2 pi/lambdo_0 x n x 2d = m (integer) 2 pi, so 1/lambda_0=m/(2 n d), m= 1, 2
Because first reflection has a negative sign, and all remaining ones have positive phase at interfaces defines the zero reflection condition as per last day.
Noted that the peak transmission are associated with modes or poles of the response of the thin film, referring to the r_12 formula on slide 4. The real part of the denominator of t_12 goes through zero, so it would be infinite except for the fact that the imaginary part is nonzero. The imaginary part is what determines the width of the resonances.
Condition (equation) for zero reflection?

6. Vary e for fixed d=600 nm d=600 nm e=13 d=600 nm e=4 d=600 nm e=2.4
Note the width of the resonances gets wider the smaller is n…why?
Light trapped in the film, bouncing back and forth, will stay in the film longer, the higher the reflection coefficient is. Longer time bouncing back and forth in time, corresponds to a wider bandwidth in frequency (1/lambda) space. The reflectivity is from the Fresnel reflection coefficient, so goes as (n_1-n_2)/(n_1+n_2), so gets bigger the larger the refractive index contrast.

7. Moving forward
Cast our result in the form of a matrix equation that yields the reflected and transmitted wave amplitudes (the out-going wave amplitudes) when the incident wave amplitude from the left hand medium is specified.

8. Moving forward
Cast our result in the form of a matrix equation that yields the reflected and transmitted wave amplitudes (the out-going wave amplitudes) when the incident wave amplitude from the left hand medium is specified.
Ask them to “fill in the matrix” when writing the input output equation with a blank S matrix before showing result

9. Moving forward
If you had an incident wave from the right hand side, how might you express the corresponding matrix representation of the out-going field amplitudes?

10. Moving forward
Can you combine these to generate a matrix equation that yields the outgoing waves for the general case when you have two in-coming waves, one from each side?

11. The S and M matricies
Would this S matrix help you easily solve for the overall reflected and transmitted fields if you had multiple dielectric layers up against each other?
What would the ideal “transfer matrix” be that would allow you to get the overall transmission simply by multiplying matrices for each boundary?

12. The S and M matricies
Can you obtain this desired matrix, call it the M matrix, from the S matrix?

13. The S and M matricies
What is then the algorithm for obtaining the overall transmission and reflection amplitude for plane waves incident on an arbitrary number of dielectric films arranged in a planar stack?

14. Let’s play! … … n1 d1 n2 d2 n1 d1 n2 d2 n1 d1 n3 d3 n1 d1 n2 d2 n1 d1
This a very general geometry, could apply to HeNe laser cavity, semiconductor laser, anti-reflection coatings on glasses …
nlayers-1
nlayers-1

15. Last 15 minutes Q&A (open forum)
- what range of n available?
- how deal with a continuously varying n(z)?
- what about total internal reflection?