Doublet Athermalization Example Introduction and Setup
So we’ve talked about how to mechanically athermalize a doublet in this case, and we’re going to move on to an example of how to calculate for L4. So, using a commercially off the shelf achromatic doublet we can actually calculate the thermo-optic coefficient, and the thermo-optic coefficient of a particular, of a specific lens element is actually equal to the thermal expansion coefficient of that lens material minus one over the index of refraction minus one times the change in index of refraction with temperature, and it has units of one over degrees Celsius. Now, for a doublet we can calculate the individual thermo-optic coefficients of each lens element using this equation and then we can bring them together in order to determine the thermo-optic coefficient of the doublet itself.
Determining the thermo-optic coefficients of each lens element
So in this example lens element 1 is a N-BAF10 and lens element two is N-SF10. These have well-characterized parameters for each of these variables in this equation. So the thermal expansion coefficient of lens element 1 is equal to 6.18×10^-6. The reference index of refraction is 1.67003 and the change in index of refraction of lens element 1, dT, in the case we’re gonna be going from 25°C to 50°C is equal to 3.849×10^-6. For lens element 2 thermal expansion coefficient is equal to 9.40×10^-6. The index refraction is equal to 1.72828 and the change in index of refraction with temperature is equal to, ah between 25°C and 50°C is equal to -7.012×10^-10.
So now we can determine the thermo-optic coefficient of lens element 1, and that’s equal to 6.18×10^ -6, minus 1 over 1.67003 minus 1, times 3.849×10^-6. And the thermo-optic coefficient of lens element 2 is equal to 9.40×10^-6, minus 1 divided by 1.72828 minus 1, times -7.012×10^-10. Now this side of the equation simplifies to 5.745×10^-6. So you have -6, -6 here, similar magnitudes. On this side we have -6 and -10. This term is so small that it essentially can be thought of as zero because it has such a tiny contribution to the thermo-optical coefficient compared to the thermal expansion coefficient. And so we have something kind of interesting here where the dndT of lens element 2 is so small that the thermal expansion coefficient dominates whereas for lens element 1 the thermal expansion coefficient and the and the dndT term kind of each cancel each other out. And so the thermo-optic coefficient for lens element 1 is quite small at 0.435×10^-6. 1 over ° Celsius. And the thermo-optic coefficient of lens element 2 is equal to 9.40×10^-6. So even though the dndT of this lens material is tiny, you do have a significant thermo-optic coefficient, and even though the dndT of lens element 1 is significant it cancels out the positive thermal expansion of the lens material.
Visualizing the focal length impact of thermo-optic coefficients
So, now I’d like to actually visualize what’s going on here. So, on this axis if we plot the change in focal length over the original focal length, you can think of this kind of like strain, also Delta L over L naught. And this would be 0, this would be 200, this would be 400, 600 and 800, all times 10^-6. And this would be -200, -400, -600, -800. And if this is temperature, on this is axis, this is 25°C, this is 50°C. We can plot these thermo-optic coefficients by determining the slope of a line on this graph. So the displacement, or the focal length change if you will, of lens element 2 will be positive and the slope right here is Beta 2. So at 25° C, that’s our reference temperature, that’s zero. At 50°C, you have close to a 200×10^-6 change because remember this is 9.4×10 ^-6 per °C. So a 25°C change will result in a focal shift. Now Beta 1, this is very very small and actually hugs the zero line and so this would be lens element 2, and this would be lens element 1, and there is a slope there it is quite small, but that slope is Beta 1. And so, both of these thermo-optic coefficients will come together to create the thermo-optic coefficient of the doublet.
Calculating the thermo-optic coefficient of the doublet
And the equation to combine these two is quite simple, it’s the thermo-optic coefficient of the doublet is equal to the focal length of the doublet divided by the focal length of element one, times the thermo-optic coefficient of element plus the focal length of the length of doublet divided by the focal length of element 2, times the thermo-optic coefficient of lens element 2. Now this equation is actually quite similar, it’s of a very similar form, not exactly, but a very similar form to the way we expanded the thermal expansion coefficients and combined the thermal expansion coefficients and the length of the two housing materials to athermalize the system. So for this particular lens, or this particular doublet we’re looking at, the focal length of the doublet is equal to 43.566 mm, focal length of lens element 1 is equal to 21.413 mm, and the focal length of element 2 is equal to 33.8093 mm. We can use those in this equation and what we calculated previously to determine the thermo-optic coefficient of the doublet. So bringing all of it together the thermo-optic coefficient of the doublet is equal to 43.5606 divided by 21.4133, times 0.435×10-6, which is that, plus 43.5606 divided by -33.8093, times 9.40×10-6. This term goes to 2.03. This term goes to 1.29. Multiplying out, the thermo-optic coefficient is equal to 0.885×10^-6 minus 12.11×10^-6. Therefore the thermo-optic coefficient of the doublet is equal to -11.22×10-6, 1 over ° Celsius. This is very interesting because this is negative. In fact, the thermo-optic coefficient of the doublet it’s not in between the thermo-optic coefficient of lens element 1 and 2, it is negative. And the reason is, the lens element that has the highest contribution to the thermal foc, the thermal focal shift, has a negative focal length and that negative focal length results in the domination of the thermo-optic shift or the thermo-optic coefficient of the doublet itself.
Negative focal lengths and athermalization with ALLVAR Alloy 30
So we don’t really have any real materials normally that have a negative thermal expansion coefficient, but our ALLVAR alloy materials does. So this is the thermal expansion of ALLVAR alloy 30, where the slope is Alpha ALLVAR. And this is the thermal expansion coefficient of aluminum. And so we can tune the, any thermo-optic coefficient that falls in between this line and this line, anywhere in between, we can use a series of aluminum and ALLVAR alloy 30 to athermalize an air gap. And in this case, we can use it to athermalize the thermo-optic coefficient of this doublet.
So we know that the length of the housing is going to be equal to the focal length of the doublet, and in this case it’s equal to 43.5606 mm, that’s already a known number. We have the thermo-optic coefficient, and we can then use that in the equation that we developed for a particular housing element, in this case Element 4, and setting it equal to the thermo-optic coefficient of the doublet minus the thermal expansion coefficient of another housing material divided by thermal expansion coefficient of Element 4 minus the thermal expansion coefficient of another housing material, all times the length of the housing which again is equal to the focal length of the doublet. Now we can set Alpha 4 to be equal to the thermal expansion coefficient of ALLVAR, and we can set the thermal expansion coefficient of aluminum equal to Alpha 3 and that will equal 24.6×10^-6. In this case it’s equal to -30×10^-6, and this is 1 over degrees Celsius, one over degrees Celsius.
Determining the length of ALLVAR Alloy 30 needed to athermalize the air gap
Now we can bring it all together. So the length, instead of Element 4 the length of the ALLVAR component is equal to -11.22×10^-6 minus the thermal expansion coefficient of aluminum 24.6×10^-6 divided by the thermal expansion coefficient of ALLVAR: -30×10^-6, minus thermal expansion coefficient of aluminum: 24.6×10^-6, all times that focal length, which is 43.566 mm. The units here will cancel, so the length of the ALLVAR to athermalize the doublet we’ve been talking about is going to be equal to 28.578 mm and Delta T in this case is between 25° C and 50°C.
Athermalize more complex systems with these formulas
But the beauty of this equation is that it doesn’t need to be confined to only a doublet. Any thermo-optic coefficient for any complex system can be plugged in to determine the length of ALLVAR needed relative to aluminum to athermalize a system. So now you can take your known tried and true aluminum SM1 tubes and you can prototype with an ALLVAR tube and athermalize air gaps, athermal doublets, and athermalize more complex lens systems.
