Ideal index of refraction for single layer AR coating

AI Thread Summary
The discussion centers on determining the ideal index of refraction for single and multiple layers of anti-reflective (AR) coatings. It emphasizes that minimum optical losses occur when the refractive index of the coating layers is the geometric mean of the adjacent materials. For a single layer matching an index of 1.52 to 1.0, the desired index is calculated as ndesired = (n1⋅n2)-2. When extending this to two layers, the second layer's refractive index should also be the geometric mean of the first layer and the substrate, leading to a more complex relationship between the indices. The conversation also touches on the implications of using materials with higher refractive indices than the substrate and how that affects phase changes and thickness calculations.
Blanchdog
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Homework Statement
Not a homework problem, just doing some review for a test. I have seen it claimed that for crown glass (ng = 1.52), the ideal index of refraction for a single layer anti-reflective coating is approximately n = 1.23, which supposedly can be calculated from Fresnel's equations. I don't understand how, since it seems to me that optimal transmission can be achieved with any index n as long as you vary the thickness d to be a quarter wavelength for destructive interference. Or actually, would that change too if n > ng, since there wouldn't be a phase shift at the coating->glass interface? For simplicity let's just consider S polarized light at normal incidence.
Relevant Equations
$$R(s) = |\frac{n_i - n_t}{n_i + n_t}|^2$$
$$T(s) = |\frac{2n_i}{n_i+n_t}|^2$$
See statement.
 
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It occurs to me in hindsight that the double interface Fresnel equation would have been more useful, but even with that it seems like d can be varied to choose any n. But that doesn't agree with what I read nor does it consider differences in phase change thanks to n1 potentially being larger than n2 (or ng).
 
Like any 'matching' operation, be it optical, electrical, acoustic, mechanical, you get minimum losses between layers if the ratio of the characteristics between layers is the geometric mean of the adjacent materials;

i.e. for your example you are trying to match 1.52 to 1.0, the minimum loss situation with a single interface layer is ndesired = (n1⋅n2)-2.

Now see if you can extend that to two layers of AR coating on the glass surface.

And then try it between two glass surfaces, each with a different n.

Have fun!

Cheers,
Tom
 
Tom.G said:
Like any 'matching' operation, be it optical, electrical, acoustic, mechanical, you get minimum losses between layers if the ratio of the characteristics between layers is the geometric mean of the adjacent materials;
Thanks, that helps a lot with understanding the math of what is going on. I've never encountered a geometric mean in relation to a 'matching' operation though, could you help me understand what the math is describing?

As an aside: For two layers of AR coating you would want the second layer to have an index of refraction that is the geometric mean of the first layer and the substrate, right? But that would change the ideal index of the first layer, since we have ##n1 = (n0n2)^{-2}## and ##n2 = (n1n3)^{-2}##. So we have two equations in two unknowns (n1 and n2), so if we substitute either equation into the other we can get an expression for the ideal n of that layer.

Not sure what you mean by between 2 glass surfaces; wouldn't the n of the AR coating just be calculated the same way?

Follow up question if you have time; would you ever use a material with ##n>n_{substrate}##, and if you did would it change the ideal thickness calculation because there would be no phase change as the light moved from the AR coating to the substrate?
 
Blanchdog said:
As an aside: For two layers of AR coating you would want the second layer to have an index of refraction that is the geometric mean of the first layer and the substrate, right? But that would change the ideal index of the first layer, since we have and . So we have two equations in two unknowns (n1 and n2), so if we substitute either equation into the other we can get an expression for the ideal n of that layer.
Ok, we started with a simple example with a single layer of AR coating.
Tom.G said:
i.e. for your example you are trying to match 1.52 to 1.0, the minimum loss situation with a single interface layer is ndesired = (n1⋅n2)-2.
That gives you two interfaces; on on each side of the AR coating.

With two AR layers you have three interfaces to handle. Using the same n values of 1.0 and 1.52 and three interfaces, the ratio of the n's of the two layers would be: ndesired = (n1⋅n2)-3 = 1.15.

That makes the n seen by the light path 1.0 -> 1.15 -> 1.32 -> 1.52.

Wikipedia has a page on the Fresnel equations that shows how to calculate reflectance from an interface between two materials of different refractive indices.
https://en.wikipedia.org/wiki/Fresnel_equations

Try playing with it for different ratios of refractive indices and different number of layers to see how the total reflectance varies.

Blanchdog said:
Not sure what you mean by between 2 glass surfaces; wouldn't the n of the AR coating just be calculated the same way?
That depends on the definition you use for "calculated the same way."

Try it for hypothetical materials with n of 1.25 and 1.6.
What do you get for n of the intervening AR layer?

Cheers,
Tom
 
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