> **Note:** this is a very old page written way back in 2023. Since then, we've actually realized that this is all **completely wrong**. The math is correct, the physics isn't, and geometric methods, while suitable in optics, does not work very well in the microwave regime. > > However, this page has historical significance in that it was written very shortly after the Project Elara team was formed at RPI (in September of 2023); previously, Project Elara was a _de facto_ one-person project and the formation of a team was what laid the groundwork for all the successes of the Project that followed. Thus, it has been preserved. --- We wish to determine the optimal optical cavity shape to be able to serve as a focuser for a maser beam using the methods of geometric optics. What follows is an (incorrect but interesting) derivation. Consider a beam of light that makes an angle $\theta$ with the horizontal. Consider that beam hitting a sloped wall with an angle of declination of $\phi$. Due to Snell's law, the angle the incoming light beam makes with the wall normal is the same as the angle the outgoing light beam makes with the wall normal: ![[angles-diagram.png]] We aim to solve for $\beta$, which is the angle made by the next light beam with the horizontal. To do so, however, we must first calculate the other angles, using the simple geometric fact that $\angle a + \angle b + \angle c = \pi$. First, we calculate $\gamma$: $ \left(\frac{\pi}{2} - \theta \right) + \gamma + \left(\frac{\pi}{2} - \phi \right) = \pi $ $ \gamma = \theta + \phi $ Then, we calculate $\alpha$: $ \alpha = \frac{\pi}{2} - \gamma $ Then, we calculate $\omega$: $ \omega + \theta + \alpha = \pi $ $ \omega = \frac{\pi}{2} + \phi $ Finally, we can calculate $\beta$: $ \beta + \alpha + (\pi - \omega) = \pi $ $ \beta = \theta + 2\phi $ We can now say that $\beta = \theta_{i + 1}$, as it would be the value of $\theta$ for the next bounce, and the whole setup would repeat. From there, we obtain the equation: $ \theta_{i + 1} = \theta_i + 2\phi_i $ However, this equation continually increases, whereas what we want is for $\theta_n = 0$ as we want planar waves. Thus, we add a small correction: $ \theta_{i + 1} = \theta_i + 2\phi_i $ $ \phi_{i + 1} = \phi_i - i \epsilon $ where $\epsilon$ is a suitably small angle. Using this, we find that the resulting curve of $\phi(x)$ is a curve whose derivative is a straight line, and therefore: $ f(x) = ax^2 $ Using $f(10) = 6$ obtained from the equation, we solve for the value $a = \frac{50}{3}$. Therefore, the mathematically ideal surface is a surface of revolution about the x-axis of the curve: $ f(x) = \frac{3}{50} x^2 $ ![[focusing-chamber-result-geometry.png]]