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I'm trying to understand intuitively why the peak reflected light from a thin-film decreases in wavelength with increasing angles. To me, it seems it should be the opposite. I know the peak reflectance is given by the equation $2nd\cos(\beta)=(m-1/2)\lambda$, where beta is the angle of the refracted ray in the film. Increasing angle of incidence, will decrease beta from 1, which causes the decrease in peak wavelength.

However, I'm trying to understand this from the perspective of how the optical path length in the film changes. When the angle of incidence increases, then the optical path length will also increase. Shouldn't this mean that the peak wavelength would increase? For the bottom reflected wave to be in phase with the top reflected wave (shifted by $\lambda/2$), the optical path difference must equal $\lambda/4$.

If we imagine the thin-film to be 100 nm thick, refractive index of 2, originally it will have a peak at reflectance of $100 \times 2 \times 4=800$ nm. When we increase incidence, say the distance is now 120 nm, thus $120 \times 2 \times 4 = 960$ nm.

This is clearly wrong, because the correct equation tells me wavelength will decrease. In the figure below, you can see how peak moves to shorter wavelength with increasing angle of incidence (from S Kinoshita et al 2008 Rep. Prog. Phys.71 07640) But I don't understand why my reasoning gives the wrong answer? enter image description here

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  • $\begingroup$ Peak wavelength does increase as angle increases, for the reason you said. Sounds like there is a problem with the equation. $\endgroup$
    – mmesser314
    Jun 27, 2020 at 14:49
  • $\begingroup$ increasing the angle of incidence increases the angle in the film so it will be greater than $\beta$. Beta is the angle for the peak reflectance. $\endgroup$
    – Peter
    Jun 27, 2020 at 15:02

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For anyone who might have the same question, I've figured out the answer. The equation is right - increasing angle of incidence does shift the color towards smaller wavelengths. Where my logic goes wrong is because it is not the absolute optical path of the reflected wave from the bottom of the thin-film that matters, but the optical path difference between the top reflected wave and the bottom reflected wave.

As the angle of incidence increases, this difference decreases. If you do not believe this, check the math on the wikipedia page on thin-films (https://en.wikipedia.org/wiki/Thin-film_interference, I needed to).

There are many statements out there that state the opposite. For example, in the intro to the wikipedia article linked above itself, it says "For any certain thickness, the color will shift from a shorter to a longer wavelength as the angle changes from normal to oblique". This is INCORRECT. In the very same article, they derive the equations that show that this is incorrect.

However, it might sometimes appear that this is what happens. A soap bubble is blue at the normal angle, then changes to red. The reason for this, is that there are several orders of interference (m in the equation). So, as the first order interference blue peaks shifts to smaller wavelengths and out of the visible spectrum, the second order peak becomes visible. This second order peak was previously in the infra-red, so it has shifted to shorter wavelengths too, according to the equation. Thus, the correct statement is "For any certain thickness, the color will shift from a LONGER to a SHORTER wavelength as the angle changes from normal to oblique".

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