Let's say I have a thin lens model of an optical system. When I have a ray that is parallel to the optical axis, the situation is quite standard - the ray refracts and passes the focal point f (see my bad drawing).

refraction of a ray on a thin lens

From the triangle in the picture, I can calculate the angle $\beta$ by using the formula $\tan(\beta) = y/f$ and so $\beta = \arctan(y/f)$. But what if my ray is not parallel with the optical axis? How do I calculate the angle of the refracted ray with the opt. axis $\beta'$?

refraction of a non parallel ray on a thin lens

I thought the ray might obey simply $\beta' = \beta + \delta$ = $\arctan(y/f) + \delta$, e.g. angle $\beta'$ could be calculated by simply adding the angle a parallel ray produces when refracted on a lens $\beta$ and an angle of deviation from being parallel with the optical axis $\delta$. On the other hand, I am not sure this approach is right. All in all, I am interested in a solution that does not involve the paraxial approximation (notice I use $\tan()$ in my equations) and I would like to know the following. How does one calculate the angle of refracting rays that are not parallel with the optical axis, in the thin lens model approximation?


4 Answers 4


You can use almost the same equation as in the paraxial approximation. Just use slopes instead of angles.

Instead of $\theta'=\theta-\frac{y}{f}$, just use $m'=m-\frac{y}{f}$

No trig functions needed.

Notice this is equivalent to $\tan{\theta'}=\tan{\theta}-\frac{y}{f}$, which is equivalent to $\theta'=\arctan{(\tan{\theta}-\frac{y}{f})}$

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 1, 2022 at 17:37
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    $\begingroup$ Way to go Stephen, this is even much simpler than my explanation. Thank you. $\endgroup$
    – Martin
    Jun 2, 2022 at 9:46
  • $\begingroup$ Nice trick Stephen and interesting insight :) $\endgroup$ Jul 22, 2022 at 8:01

I realized that a ray passing through the center of the lens (let's call it ray A) does not deviate from its path. And if another ray (ray B) comes in the lens with the same angle as ray A, but does not pass the center of the lens, it has to cross ray A at the back focal plane of the lens. I drew the situation on a graph.

Ray diagram with ray passing the center of the lens and a ray parallel to it

Here, we can calculate the variable $x$ by noticing the following orange triangle:

orange triangle

From here, $\tan(\delta) = x/f$ and hence $x = f\tan(\delta)$. Next, we can notice another triangle, marked in blue. This one actually contains the angle $\beta'$ that we are interested in:

blue triangle

From here, $\tan(\beta') = \frac{x+y}{f}$. The rest is just simple algebra. $\beta' = \arctan(\frac{x}{f} + \frac{y}{f}) = \arctan(\frac{f\tan(\delta)}{f} + \frac{y}{f}) = \arctan(\tan(\delta) + \frac{y}{f})$.

All in all, when tracing a ray passing through a thin lens without paraxial approximation, I think its angle with the optical axis after refraction will be $\beta' = \arctan(\tan(\delta) + \frac{y}{f})$, where $y$ is the point measured from the center of the lens where the ray hits the lens, $f$ is the focal point of the lens and $\delta$ is the angle of the ray coming to the lens, measured from the optical axis.


Thin lenses produce a linear change in the slope of the light: $\Delta m = -\frac{h}{f}$. Now, we can relate the slope to the angle $m = \tan \theta$ to get:

$\tan \theta' = \tan \theta - \frac{h}{f}$.

Why is the change in slope linear?

Let's say our ray hits the lens at height $h$ and has slope $m$.

Consider a parallel ray that goes through the axial-center of the lens: its slope remains unchanged after the lens $m'_C=m$, so by the time it makes it to the focal plane its height will be $= m'_C\cdot\Delta x=m\cdot f$.

Our original ray will meet with this central ray at the exact same height in the focal plane $y' = m\cdot f$. Thus, we conclude its slope after the lens will be $m' = \frac{\Delta y}{\Delta x} = \frac{m\cdot f-h}{f} = m - \frac{h}{f}$.


For a thin lens, in the paraxial approximation, the ray angle after the lens is given by nu', where

nu' = nu - y * phi,

phi is the power of the lens (or reciprocal of the focal length), and nu is the angle of the ray before the lens.

In this sign convention, rays have a negative nu angle if they are running down to the right. Note that this is a paraxial description, so there are no tangents involved. The angle nu is just radians. So, for example, if the incoming ray is horizontal, the outgoing angle will just be -y/f, where f is the focal length.

The book by Warren Smith, Modern Optical Engineering, has a description of the paraxial ray trace. Just be aware of possibly different sign conventions.

  • $\begingroup$ Thank you JB2 for your answer! Unfortunately, I should have explicitly stated that I am not looking for the solution in the paraxial approximation as I already know how to solve this problem (I have edited my question to be crystal clear). Instead, as you can see, I am trying to derive the angle $\beta'$ in the general case. Anyway, thank you for your time and effort :) $\endgroup$
    – Martin
    Jan 27, 2022 at 15:54
  • $\begingroup$ Sorry, I assumed when you had a thin lens you were interested in the paraxial approximation. People don't often combine thin lenses with the general ray trace equations. If you are interested in the general ray trace equations, then the book by Smith covers that. If you are really interested in thin lenses, then look up this article: Sweatt, W. C. Describing holographic optical elements as lenses. J. Opt. Soc. Am. 67(6), 803 (1977). Sweatt describes a model for a thin lens that approximates a hologram. $\endgroup$
    – JB2
    Jan 27, 2022 at 16:31
  • $\begingroup$ The hologram has specific dispersive properties that don't apply for your case, but I suspect he covers the ray trace for such a device. You could also just use Smith's equations for a general ray trace and let the refractive index go to a very high number. This is equivalent to making the lens very thin. $\endgroup$
    – JB2
    Jan 27, 2022 at 16:31
  • $\begingroup$ Thank you for your suggestions JB2 again! I checked the article but I found it a bit off topic, but using the Smith's equations for a general ray trace with high refractive index could actually work. However fortunately, I was finally able to derive the result by myself and I will try to give a precise answer to my question. Thank you very much again! :) $\endgroup$
    – Martin
    Jan 28, 2022 at 8:26

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