3
$\begingroup$

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?


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?

$\endgroup$
3
  • $\begingroup$ Has anyone even seen this formula before? $\endgroup$ Jan 8, 2022 at 15:18
  • 1
    $\begingroup$ Perhaps it is significant that according to alluxa.com/optical-filter-specs/… the formula only holds for 'relatively small angles of incidence' $\endgroup$
    – Martin CR
    Jan 9, 2022 at 19:58
  • 2
    $\begingroup$ @MartinCR Yes I was starting to think that also, the problem is that is not the only website using that formula, hyperphysics uses it also, formulas don't just magically appear. Someone derived this formula and it is probably sitting within a book on optics somewhere, but I can't find it, so I'm stuck. $\endgroup$ Jan 9, 2022 at 23:28

3 Answers 3

7
+200
$\begingroup$

Interference filter 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 $$k_1\sin \theta_1 = k_2\sin \theta_2 \qquad \Longleftrightarrow \qquad \frac{\sin \theta_2}{\sin \theta_1}=\frac{k_1}{k_2}=\frac{n_1}{n_2}$$

$\endgroup$
2
  • 1
    $\begingroup$ Brilliant answer (not least for the typesetting!) $\endgroup$
    – Martin CR
    Jan 10, 2022 at 6:58
  • 3
    $\begingroup$ My pleasure! Anecdote: I just took apart an old LCD projector during christmas holidays, and I explained to my 6-year-old daughter why the contained interference filters change color when you tilt them. Of course without the refraction stuff, and by grossly simplifying the interference principle, which would both have been above her head. But she got the principle and that proves the worth of good explanations. $\endgroup$
    – oliver
    Jan 10, 2022 at 8:39
3
$\begingroup$

Consider a ray impinging on the mirror on the right. Some of it is transmitted, and some is reflected, bounces off the left mirror, reaches the right mirror again, and itself is transmitted. The wavelength transmitted by the interference filter is such that the optical path difference between these two rays at infinity is equal to one wavelength so that they constructively interfere.

This optical path difference can be calculated geometrically in terms of the angle $\beta$ in the spacer, its thickness $d$ and the refractive index $n$ as is done e.g. here. The result is

$$x=\frac{2nd}{\cos(\beta)}-2nd\tan(\beta)\sin(\beta)=2nd\cos(\beta)$$

and, requiring that $x=\lambda$, we find

$$d=\frac{\lambda}{2n\cos(\beta)}$$

which is what is given on the Hyperphysics website. From this we also find that the wavelength at normal incidence is $\lambda_0=2nd$ so it can be rewritten as $\lambda=\lambda_0\cos(\beta).$ Finally, Snell's law states that $n\sin(\beta)=\sin(\alpha)$ where $\alpha$ is the angle of incidence so we get

$$\lambda=\lambda_0\cos(\beta)=\lambda_0\sqrt{1-\sin^2(\beta)}=\lambda_0\sqrt{1-\frac{\sin^2(\alpha)}{n^2}}.$$

The reason this only holds for small angles of incidence is that the mirrors themselves are usually made up of quarter wavelength layers so if $\lambda$ is significantly different from $\lambda_0$ or the ray does not pass through in a sufficiently straight line they will no longer reflect properly.

$\endgroup$
1
$\begingroup$

A Remark (not a demonstration):

Same equation by analogy (the speed (v) of light in the medium is replaced by its speed in vacuum but in a moving reference frame):

Assuming that an observer is at the origin of the coordinates of the frame of reference K' and that the latter moves with respect to K parallel to the x-axis, i.e. y=y', z=z'

At the time t=0, the origins of the two reference frames are merged. it follows that in the reference frame K the coordinates of the observer are $x=vt, y=z=0$. In the K reference frame, we have from the LT for coordinates :

From the LT:$ \;\;R'^{2}=\gamma^{2}[(x-vt)^{2}+\gamma^{-2}(y^{2}+z^{2})]$

$x^{2}+y^{2}=R^{2}\sin^{2}(\theta)$

$R^{*}=\gamma^{-1} R'=R\sqrt{1-\beta^{2}\sin^{2}(\theta)}=R\sqrt{1-\frac{\sin^{2}(\theta)}{n^{2}}} $

if we pose that $\;R=\lambda_{0} \;\;, R^{*}=\lambda$, we have:

$\lambda=\lambda_{0}\sqrt{1-\frac{\sin^{2}(\theta)}{n^{2}}} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.