Question

A jewelry maker has asked your glass studio to produce a sheet of dichroic glass that will appear red for transmitted light and blue for reflected light. You decide that "red light" should be centered at 656 nm and that "blue light" is 492 nm. If you use a MgF₂ coating (n = 1.39), how thick should the coating be?
nm

Answer 1

Given that refractive index of MgF2 coating is, n = 1.39.

Now as metioned in the problem, this sort of scheme is usually used for antireflection coatings using phase cancellation, but can be used for frequency-specific reinforcement as well. For reinforcement, the phase shift of the ray reflected from the MgF-glass boundary should be equal to (or an integer multiple of 2pi different from) that of the ray reflected from the air-MgF boundary. Both reflections, being of the type where the ray travels through a lower-N medium to a boundary with a higher-N medium, produce a phase shift of pi. For the two reflected rays to be in phase, the additional two-way path length of the coating, 2T, must be an integer mult. of ?, thus T must be an integer mult. of ?/2.

Therefore -

With n = 1.39, ? in MgF = 492 / 1.39 nm.

So, thickness of the coating = (0.5 x 492) / 1.39 nm = 177 nm.