# Why does peak wavelength decrease with angle of incidence in thin-films?

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?

• Peak wavelength does increase as angle increases, for the reason you said. Sounds like there is a problem with the equation. Commented Jun 27, 2020 at 14:49
• 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. Commented Jun 27, 2020 at 15:02