PHYS 408 Applied Optics (Lecture 10) JAN-APRIL 2016 EDITION JEFF YOUNG AMPEL RM 113

Quick review of key points from last lecture S and M matricies 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 matricies are straight forward to figure out, and the associated M matricies 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 matricies sequentially to obtain M net for the entire structure, and then using linear algebra to either solve for r and t, or using the transformation properties from M net to S net.

Bottom line Consider M net as M, then S net (A,B,C,D) yields t 02,t 20,r 02,r 20 Or don’t be lazy, and just solve for and from

Can anyone think of another way to circumvent transforming from M net to S net ? What prevents you from, once you find M net, putting in values for and just multiplying it by M net to get the transmission? Would this help?

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?

Internal Field Distributions (d 2 infinite) n1n1 d1d1 n2n2 d2d2 z 0

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

Generalize n1n1 d1d1 n1n1 d1d1 n2n2 d2d2 n1n1 d1d1 n2n2 d2d2 n3n3 d3d3 n1n1 d1d1 n2n2 d2d2 … … nlayers-1

Uniform periodic multilayer stack n1n1 d1d1 n2n2 d2d2 n1n1 d1d1 n2n2 d2d2 n1n1 d1d1 n2n2 d2d2 n1n1 d1d1 n2n2 d2d2 ……

Bragg reflection n 1 =1.3; n 2 =1.4; n 3 =n 2 d 1 =400 nm; d 2 =200 nm; d 2 =d 3 21 periods

Bragg reflection n 1 =1.3; n 2 =1.4; n 3 =n 2 d 1 =580 nm; d 2 =20 nm; d 2 =d 3 21 periods

Bragg reflection n 1 =1.3; n 2 =1.4; n 3 =n 2 d 1 =580 nm; d 2 =20 nm; d 2 =d periods

Bragg reflection n 1 =2; n 2 =sqrt(12); n 3 =n 2 d 1 =300 nm; d 2 =173 nm; d 2 =d 3 10 periods

Add a “defect” n1n1 d1d1 n1n1 d1d1 n2n2 d2d2 n1n1 d1d1 n2n2 d2d2 n3n3 d3d3 n1n1 d1d1 n2n2 d2d2 ……

What is this? n 1 =2; n 2 =sqrt(12); n 3 =4 d 1 =300 nm; d 2 =173 nm; d 3 =0.15*d 2 10 periods

And this? n 1 =2; n 2 =sqrt(12); n 3 =4 d 1 =300 nm; d 2 =173 nm; d 3 =2*d 2 10 periods

Cavity Modes! n 1 =2; n 2 =sqrt(12); n 3 =4 d 1 =300 nm; d 2 =173 nm; d 3 =10*d 2 10 periods

Cavity Modes! n 1 =2; n 2 =sqrt(12); n 3 =4 d 1 =300 nm; d 2 =173 nm; d 3 =10*d 2 10 periods SYMMETERIZED