I want to know the relation between the effective refractive index and the thickness of film deposited on substrate.
Dear Yashashwa, Given the complex refractive index of substrate (a+ib), when you apply dielectric coating layer of refractive index N and thickness d, the effective refractive index Y is given by [(a+ib) cos(delta) + i N sin(delta)]/[cos(delta) + (a+ib)(i sin(delta) /N)] as function of delta=2 pi N d/lambda, where lambda is the wavelength of light and d is thickness of coating layer. Trajectory of Y on the complex plane is, therefore, a circle in this case. Attached please find an example of substrate refractive index (1.9,0) and coting layer MgF2 with N=1.38. Starting from (1.9,0), as thickness of coating layer increases, the trajectory goes round on the lower semicircle toward the imaginary axis passing through (1,0). When the thickness is equal to lambda/4N, which is so called quarter wavelength layer, the effective refractive index becomes (1.0), which is equal to that of air. Shigeo
Dear Yashashwa, Given the complex refractive index of substrate (a+ib), when you apply dielectric coating layer of refractive index N and thickness d, the effective refractive index Y is given by [(a+ib) cos(delta) + i N sin(delta)]/[cos(delta) + (a+ib)(i sin(delta) /N)] as function of delta=2 pi N d/lambda, where lambda is the wavelength of light and d is thickness of coating layer. Trajectory of Y on the complex plane is, therefore, a circle in this case. Attached please find an example of substrate refractive index (1.9,0) and coting layer MgF2 with N=1.38. Starting from (1.9,0), as thickness of coating layer increases, the trajectory goes round on the lower semicircle toward the imaginary axis passing through (1,0). When the thickness is equal to lambda/4N, which is so called quarter wavelength layer, the effective refractive index becomes (1.0), which is equal to that of air. Shigeo
Thank You Dr. Kubota for you response. It really helped me a lot.
However I am coating my chip with varying thickness of very thin (few nano meters) of SiO2 to study its plasmonic behaviour with varying refractive index but I am mainly considered with the real part of the refractive index, so is there a simple expression for that? And yes the wavelength of light is much larger than the thickness of the SiO2 layer.
Dear Yashashwa, Assuming substrate material is Si, (n=3.882+i0.019 at 633nm), the trajectory of Y is plotted on complex plane as a function of delta=2 pi n'd/lambda, where n'=1.46 at 550nm. I also plotted Re.Y as a function of SiO2 thickness d in nm, which is approximated by a parabolic function such as 3.8849+0.00456d-0.003982d^2. Of course you can improve approximation using more precise input data. Please see attached file updated. Shigeo
Dear Yashashwa, I should comment about your concerns about applicability of approach using admittance diagram to very thin layer in research of plasmonic behavior. In Fig 8.9 of A. Macleod’s Thin film optical filters, there is an example applying it for 0.002 wavelength water layer on Silver. Shigeo
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