What happens to the bandpass of a dichroic filter when it is tilted?

[Drakkith]

Say you have a dichroic filter with a bandwidth of 10nm centered at 500nm, so it will pass 495nm - 505nm light through.
In an optical system the light is generally focused as a cone through the filter, which means that most of the light, if not all, is entering the filter at an oblique angle. What effect does this have on the light passing through the filter? I assume that there is a small change in the bandpass for light that enters at an oblique angle. Is that correct?
And the same thing happens if you tilt it too?
Does the bandwidth stay relatively the same if you tilt the entire filter, with only the upper and lower bandpass frequencies going up or down, or does the bandwidth change too?

 

[Redbelly98]

Just thinking this through based on my own knowledge of optics, though it has been 9 years since I actively worked in this area.

These filters are typically made from coating the glass with multiple layers of materials with different refractive index, which sets up multiple reflections within those layers. The resulting interference then transmits the desired wavelengths and reflects the unwanted wavelengths. Sorry if you already knew that, but I want to be sure we're on the same page in our understanding.

As the angle deviates from normal, the beam must go through more material in the optical coating layers, so I would expect the bandpass wavelengths to increase by a factor of (1/cosθ), where θ is the angle from the normal in the coating material. The actual angle of incidence of the beam in air will be at an angle larger than θ.

That (1/cosθ) factor would apply to both the 495 nm and 505 nm wavelengths, and also to the 10 nm bandwidth.

That being said, if the angle becomes very large then the usual reflection & transmission coefficients will deviate significantly from their normal-incidence values, which I expect will mess with transmission inside the bandwidth and the reflection outside the bandwidth.

Hope that helps -- we can try to get more quantitative if you like, but hopefully that gives some idea. I used to work with lasers a lot in a former career, but we pretty much used lenses, mirrors, etc. at the design incidence angle -- I never really played around to see what happened if we got away from that.

 

[mfb]

As those different layers have a different refractive index, the factor cos(θ) would be different for different layers. This could change the bandwidth and make the filter worse in general. I don't know how significant that effect is.

 

[Drakkith]

Ok, with an F/4 mirror of 254 mm I'm getting an angle of 7.13 degrees from normal for the light at the edge of the converging light cone. How would I go about calculating the change? I'm assuming I would need to know the refractive index of all those different layers?

 

[Redbelly98]

mfb brings up a good point about the refractive index being different for the coating layers. But we can at least come up an estimate. After a little searching, it looks like the coatings can have n of 1.4 to 2.3 or so. So that 7 degree angle becomes 3 to 5 degrees (applying Snell's Law to 7 degrees). 1/cosθ is then 1.0014 to 1.0038 -- a pretty small effect. At 500 nm, that's a shift of 0.7 to 1.9 nm.

p.s. I googled optical coating index and found useful info at:
http://en.wikipedia.org/wiki/Optical_coating#High-reflection_coatings
http://www.edmundoptics.com/learning-and-support/technical/learning-center/application-notes/optics/an-introduction-to-optical-coatings/
http://photonics.com/edu/Handbook.aspx?SID=29&AID=25494

 

[Drakkith]

Oh wow, that's a pretty big change for some filters. Thanks Redbelly!

 

[Cthugha]

Actually dichroic filters are a mean pit trap because they behave exactly the opposite way of what one naively assumes. In fact, the angle-dependence of the transmission wavelength behaves roughly like \lambda(\theta)=\lambda_0 \sqrt{1-(\frac{n_0}{n_{eff}})^2 \sin^2(\theta)}
(with the n's being the refractive indices of the surrounding material and the effective refractive index of the coating) and the transmission spectrum will shift to shorter wavelengths as the incidence angle gets larger.

In a nutshell, the explanation for this counterintuitive effect is that you can split the k-vector of the incoming wave into a part parallel to the filter surface and one part perpendicular to it. The resonance condition depends on the accumulated phase and is roughly given by the condition that the k-vector-component perpendicular to the filter surface times the thickness of the layer should be a multiple of pi. The k-vector-component parallel to the surface does not matter in that respect. Therefore you need a larger total k-vector to get the same k-vector component perpendicular to the surface and as energy scales with k, you get a shift to higher energies/smaller wavelengths.

From my experience it is possible to shift the wavelength 4-5 nm for a central wavelength in the 700-800 nm range. However, the transmission spectrum may change a bit when one really goes to angles larger than 20-30 degrees.

See for example the following data for dichroic filters under normal incidence and incidence under different angles to get some feeling for the magnitude of the shift:
http://www.thorlabs.us/newgrouppage9.cfm?objectgroup_id=986

 

[Drakkith]

Interesting. The bandpass isn't just shifted in some cases, it's also compressed in some too.