What is the angle of a ray passing through a thin lens?

Question

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).

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 $\beta'$?

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?

Answer 1

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 $\theta'=\arctan{(\tan{\theta}-\frac{y}{f})}$

Answer 2

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.

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

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.

Answer 3

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 $m' = \frac{\Delta y}{\Delta x} = \frac{m\cdot f-h}{f} = m - \frac{h}{f}$.

Answer 4

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.