1. Multiple Beam Interference at Dielectric Interfaces n1n1 n2n2 n3n3 If the reflectivity is high, multiple reflections can not be ignored Things can get quite complicated for even more layers! …….. n1n1 nini njnj nNnN Let’s develop a matrix formalism (again!) to handle the phase shifts resulting from a general case.

2. Phase shifts at an interface …….. i j E ri E rj E’ ri E’ rj E li E’ li E lj E’ lj i, j: i-th and j-th layers r, l: fields propagating in the right and left directions Primed, unprimed: right and left side of a given layer Use Stokes relations to simplify

3. Transition matrices Interface transition matrix Now, let’s determine the phase change upon traversing a layer (we will restrict ourselves to normal incidence) phase change upon moving through a j-th layer with index n j is k*d=n j (2  / ) d  j Layer propagation matrix

4. Synthesis of it all …….. n1n1 nini njnj nNnN Notice that the final region (N th ) has no component moving to the left:

5. Application I:Anti-reflection Coating n 1 < n 2 n 2 < n 3 n 1 = 1n 2 = ?n 3 = 1.4    /2 Reminder: 180 degree phase shift upon reflection off an interface with a more dense medium. air glass Goal: We want to have a coating of certain thickness so that the destructive interferences between the two rays will give the minimum reflectance for a target wavelength, typically in the middle of visible wavelengths (~550nm). d Why do we need n 1 < n 2 < n 3 ? The reflectivity of two interfaces should match to provide the maximum cancellation between two light fields of comparable strength. Even if these conditions are satisfied, the reflectance for other wavelengths will be non-zero. Let’s evaluate how much better we can do compared to the regular 4% reflection in the absence of the AR coating. -> Use matrix formalism

6. Anti-reflection Coating n 1 < n 2 n 2 < n 3 n1n1 n2n2 n3n3    /2 air glass d Stack matrix for AR coating

7. Anti-reflection Coating n 1 < n 2 n 2 < n 3 n1n1 n2n2 n3n3    /2 air glass d If we require r=0 for  =  /2 (see above), then e i  /2 =i, e -i  /2 =-i. In practice, MgF 2 with n 2 =1.38 is popularly used for its durability. Minimum reflectance is 1%, not 0%.

8. Application II: Enhanced Reflection Coating   /2     ……. nHnH nLnL nLnL nLnL nHnH nHnH 33 33 55 55 n H > n L Constructive interferences from multiple layers of coating can give near 100% reflectance! Why is it better to have multiple layers? Let’s find out using matrix formalism.

9. Enhance Reflection Coating For a pair of two layers (N=1), We can rewrite S N=1 as Then for two pairs of layers (N=2) For N pairs of layers

10. Enhance Reflection Coating For N pairs of layers Let’s simplify Therefore, R can approach 100% for a very large number of N. A laser cavity mirror with 7 layers of ZnS (n H =2.32) and MgF 2 (n L =1.38) giving a peak reflectance of 99.7%.

11. More is Better n 1 = 1 and n 2 = 3.556

12. More is Better n 1 = 3.45 and n 2 = 3.55

13. DBRs and DFBs