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As I could understand, we can trace rays through optical elements using the ray transfer matrices . However, the references I could find online seems like it considers the angles to be small (Sin(t) = t). For example:

enter image description here

What should I change to make the transfer working for general case where the angle is not small ? should replace all the angles with its tangent?

Edit:

I made a small calculation for the propagation matrix:

$$\begin{pmatrix}1& d\\ 0& 1\end{pmatrix}$$ It gives: $$y_2 = y_1 + d\theta_1$$ However, the correct equation based on geometric principles should be: $$y_2 = y_1 + d\tan \theta_1$$ So, I supposed that in general case we should use $\tan \theta$ instead of $\theta$. Any confirmation?

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This won't work. See for example the matrix for going from air to glass:

In the linear case, $1\cdot \theta_1 = 1.5\cdot \theta_2 \Rightarrow \theta_2 = \frac{1}{1.5} \theta_1$, meaning the matrix is $\begin{pmatrix} 1& 0 \\ 0& \frac{1}{1.5} \end{pmatrix}$. But in more general, $\theta_2 = \arcsin(\frac{1}{1.5} \sin \theta_1 ) $, so your solution of writing $\tan \theta_2 = \frac{1}{1.5} \tan \theta_1 $ isn't correct.

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  • $\begingroup$ Thanks for the edit and the answer! So, you suggested that I should replace the angle with its sin? or there is no general rule for all kind of matrices? $\endgroup$ – Humam Helfawi Nov 16 '18 at 12:58
  • $\begingroup$ There is no general rule for all kind of matrices $\endgroup$ – Ofek Gillon Nov 16 '18 at 18:02
  • $\begingroup$ @Humam Helfawi Ray transfer matrices only hold in Gaussian approximation. That is, both distance of ray from optical axis and its "vergence" (angle formed with axis) must be small. The first condition means that refraction equation is also approximated to a linear one. If these conditions are not met, and if you exactly compute the ray's path, you will find "aberrations" of various kinds: in general, rays starting from a point in object space will not converge to a point in image space. In Gaussian approximation instead this always happens (happily). $\endgroup$ – Elio Fabri Nov 17 '18 at 11:18

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