Optics – How to Derive the Angular Shift Formula for Interference Filters

Question

I came across this formula from this optical coatings website

$$\lambda_\theta=\lambda_0\left[1-\frac{\eta^0}{\eta^*}\sin^2\theta\right ]^{1/2}\tag{1}$$
Where $\lambda_\theta$ is the peak wavelength at incident angle $\theta$, $\lambda_0$ is the peak wavelength at normal incidence, $\eta^0$ is the refractive index of the incident medium and $\eta^*$ is the effective index of the filter assembly.

$(1)$ predicts the amount a wavelength is 'blue-shifted' due to the rotation of an interference filter away from normal incidence for an incident light beam.
But after lots of googling for a derivation and searching many websites such as this one, and this, I just cannot seem to find it's derivation (the websites just state the result without derivation) so I ask the question here to see if anyone knows how to derive it.

If I had to guess I would use Snell's law $\eta_1\sin\theta_1=\eta_2\sin\theta_2$ and Bragg's interference condition $N\lambda=d\sin\theta$, where $N\in \mathbb{Z}$ and $d$ is the distance between Bragg planes, but this approach doesn't seem to be getting me any closer to the desired eqn $(1)$.

Does anyone know how to derive $(1)$ or give me hints/tips towards doing so myself?

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Edit in response to comment:

A variant of this formula is given on this hyperphysics website page and looks like $$\lambda=\lambda_0 \sqrt{1-\frac{\sin^2 \alpha}{\eta^2}}\tag{2}$$ the symbols they use are defined on their webpage, but still there is no derivation for it.

Is anyone able to derive $(2)$ instead?

Answer 1

For the purpose of the question, I find it helpful to formulate the laws of refraction in terms of wave vectors $\vec k$. With it a plane wave (which we are assuming here) can be easily described by a complex exponential $$\psi(\vec r,t) = \psi_0 e^{i(\vec k\cdot \vec r-\omega t)}$$ We do not need this representation explicitly, but are rather only interested in the phase $$\varphi(\vec r,t)=\vec k\cdot \vec r-\omega t$$ Looking at planes of constant phase, we can see that the phase velocity (which is referred to by the refractive index) is in terms of the wave vector $$v=\frac{\omega}{|\vec k|}$$ For refraction between two media, the corresponding phase velocities are inversely related by the refractive indices of the media $$\frac{v_2}{v_1}=\frac{n_1}{n_2}$$ Since frequency stays the same during refraction, we can also write (we square it already for the subsequent derivation) $$\frac{|\vec k_1|^2}{|\vec k_2|^2}=\frac{n_1^2}{n_2^2}$$ Now, continuity at the refracting interface implies that a two-dimensional section of the wave travels along the interface on both sides, so the component $k_\parallel$ of the wave vector parallel to the interface is the same in both media (green arrow in the illustration above). By looking at the reflected wave vector (which has the same magnitude as the incident one) and considering the angle of incidence $\theta_1$, we can easily calculate the parallel component: $$k_\parallel = k_1\sin \theta_1$$ On the other hand, we can write the magnitude squared $|\vec k_2|^2$ componentwise $$|\vec k_2|^2=k_{2,\perp}^2+k_\parallel^2=k_{2,\perp}^2+k_1^2\sin^2 \theta_1$$ Using the above relation for the refractive indices, we get $$k_1^2=\left(k_{2,\perp}^2+k_1^2\sin^2 \theta_1\right)\frac{n_1^2}{n_2^2}$$ and if we resolve this for $k_{2,\perp}$ we obtain $$k_{2,\perp}=k_1\sqrt{\frac{n_2^2}{n_1^2}-\sin^2 \theta_1}$$ In order to investigate the interference between the refracted and reflected waves, we can equivalently look at the 2D-wave at the interface instead of the whole reflected wave, because, according to Huygens' principle, this surface wave acts as the source of the reflected (as well as the refracted) wave.

So we interfere the green path with the red path. The spatial phase at the point of interference contains $k_\parallel$ on the green as well as on the red path: $$\Delta \varphi_1 = k_\parallel\Delta h$$ $$\Delta \varphi_2 = k_{2,\perp}\cdot 2d+k_\parallel\Delta h$$ so that the relative phase difference between the two different waves amounts to $$\Delta \varphi=\Delta \varphi_2-\Delta \varphi_1=2dk_1\sqrt{\frac{n_2^2}{n_1^2}-\sin^2 \theta_1}$$ Because of $k_1=2\pi/\lambda$ and the phase difference $\Delta \varphi$ must be (a multiple of) $2\pi$ for maximum constructive interference, we get the interference condition for first order: $$\lambda=2d\sqrt{\frac{n_2^2}{n_1^2}-\sin^2 \theta_1}$$ For perpendicular incidence $\theta_1=0$ we obtain the special case $$\lambda_0=2d\frac{n_2}{n_1}$$ and thus we can finally write the general case for the incidence angle as $$\lambda=\lambda_0\sqrt{1-\frac{n_1^2}{n_2^2}\sin^2 \theta_1}$$ Note that the formula from the optical coatings website is obviously wrong in that they missed the square of the refractive indices (most likely a typo, I guess).

Note: it is also easy to prove Snell's law with the above presentation, by the way. Because we already know that $k_\parallel$ is the same for the second medium, we also have $$k_\parallel = k_2\sin \theta_2$$ and so $$k_1\sin \theta_1 = k_2\sin \theta_2 \qquad \Longleftrightarrow \qquad \frac{\sin \theta_2}{\sin \theta_1}=\frac{k_1}{k_2}=\frac{n_1}{n_2}$$