PHYS 408 Applied Optics (Lecture 10) Jan-April 2017 Edition Jeff Young AMPEL Rm 113

Quick review of key points from last lecture S and M matrices are associated with the transfer of fields across each interface, and their propagation through uniform films. The matrix elements of S for going across interfaces are obtained from the Fresnel reflection and transmission coefficients. The matrix elements of S for propagating through a uniform film include diagonal phase accumulation terms only. The S matrices are straight forward to figure out, and the associated M matrices come from transforming the S matrices using linear algebra. The net r and t for a stack of thin films is obtained by multiplying all M matrices sequentially to obtain Mnet for the entire structure, and then using linear algebra to either solve for r and t, or using the transformation properties from Mnet to Snet.

Bottom line Consider Mnet as M, then Snet(A,B,C,D) yields t02,t20,r02,r20 Or don’t be lazy, and just solve for and from

Uniform periodic multilayer stack … …

Bragg reflection n1=1.3; n2=1.4 d1=580 nm; d2=20 nm 21 periods Sketch this profile on board, and get them to figure out what condition defines this high reflectivity wavelength. - Ans: 1/lambda_0=1/(2n_ave (d1+d2)) (reflections from all periods in phase), since to a good approximation it is a uniform thin film of n=1.3 and thickness 21*600 nm.

Physics of peak with max reflectivity More generally lambda_0/n_ave~2* period of stack/m where m=1,2,3… There is a subtlty as to how one comes up with the average n (n_ave). Should you use thickness weighted n’s, or epsilons? …Try it and find out. Note that this equation is only approximate, but is remarkably handy for estimation purposes.

Change duty cycle n1=1.3; n2=1.4; n3=n2 d1=400 nm; d2=200 nm 21 periods What are differences due to change in duty cycle? (Peak R and wavelength of peak R) Bragg reflection wavelength1/lambda=1/(2n_ave d) (reflections from all periods in phase), and same equation holds for the overall thin film effect if d is replaced with the total stack thickness (smaller ripples) as per last time):

Change number of layers n1=1.3; n2=1.4; n3=n2 d1=580 nm; d2=20 nm 401 periods Despite small n1-n2, if have enough periods, r^2 very close to 1.

Change index contrast n1=2; n2=sqrt(12); n3=n2 d1=300 nm; d2=173 nm; d2=d3 10 periods Only 10 layers! Very relevant to VCSEL lasers.

Can anyone think of another way to circumvent transforming from Mnet to Snet? What prevents you from, once you find Mnet, putting in values for and just multiplying it by Mnet to get the transmission? Would this help? Don’t know what to put in for the backward reflected amplitude, so end up with system of equations. Think of numerical implementation, and simplifying it

Going “inside” the structure There is a very significant advantage to this approach. or for instance, in our anti-refection example: What does the right hand side of the following equation give you? Give them time to work on this

Internal Field Distributions (d2 infinite) z

And what can you do with all the intermediate values? You should verify this agrees with the previous result: Ie, working backwards agrees with working forward Recall

Generalize: add a defect between 2 mirrors … … nlayers-1 nlayers-1

What is this? n1=2; n2=sqrt(12); n3=4 d1=300 nm; d2=173 nm; d3=0.15*d2 10 periods Discuss relevance to the cavity experiment in the labs. Almost no light gets through either mirror on its own, but line them up, and almost all the light gets through!

And this? n1=2; n2=sqrt(12); n3=4 d1=300 nm; d2=173 nm; d3=2*d2 10 periods

Cavity Modes! n1=2; n2=sqrt(12); n3=4 d1=300 nm; d2=173 nm; d3=10*d2 10 periods

Cavity Modes! n1=2; n2=sqrt(12); n3=4 d1=300 nm; d2=173 nm; d3=10*d2 10 periods SYMMETERIZED