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In the textbook, usually the following construction (see the figure below) is used to determine the relative size of the image and the object. enter image description here

I understand that a ray parallel to the optical axis will be refracted and go through the focus, and a ray passing the vertex will continue straight ahead. The first statement is a consequence of thin lens formula and lensmaker's equation (or based their derivations). The second statement is a consequence of thin lens.

Now, in the plot, the three red rays originated from the tip of the object on the left hand side of the lens will meet at the tip of the real imagine on the right hand side of the lens. I understand this is consistent and can be shown as a result from the geometric layout of the plot and Newton's version of thin lens formula: $x_1x_2=f^2$.

Considering a thin lens consisting of two spherical surfaces, for paraxial rays, one can derive the lensmaker's equation etc. My question is: how to show that a fourth arbitrary ray (which is still paraxial but is not parallel to the axis neither go through the vertex, not shown in the plot) from the tip of the object should also go pass the tip of the image?

Maybe I have missed something obvious. However, I understand that the derivation of lensmaker's equation in a standard textbook (by explicitly using Snell's law or by using Fermat's principle) only shows how light rays emitted from a fixed point on the optical axis will all meet on another point on the optical axis. Here, the tip of the object involves a small deviation from the optical axis. It is not clearly to me, how to show these light rays are get focused onto the same image point, with higher order (therefore negligible) deviations? Many thanks!

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  • $\begingroup$ Isn't this the very definition of a lens? Are you asking for the derivation based on the actual shape of the lens? - You could turn it around and ask what shape of lens would give rise to all rays converging to the same focus. if you do not have a shape of a lens defined, then unless you accept the definition of a lens, you cannot begin to answer your question. So - which would you like? $\endgroup$
    – Floris
    Jun 6, 2016 at 2:20
  • $\begingroup$ Thanks for pointing it out. I meant the first case -- thin lens made from spherical surface. I modified the question accordingly. $\endgroup$
    – gamebm
    Jun 6, 2016 at 12:34

2 Answers 2

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One way to approach this is to use the fact that the optical path lengths of all the rays, those shown and those not shown, are all the same. The optical path length through the vacuum is equal to the actual path length. The optical path length through the glass is increased by a factor of $n$, the refractive index.

You have to make certain approximations. Notably, take the actual thickness of the lens to be zero (thin lens approximation) but account for the action of the lens by adding a segment of path length (but no offset to the position of the ray!) to represent passage through the glass. The extra thickness will be greatest at the center, and get smaller as you move away from the center, varying ... well, you have to choose. Quadratically, or spherically, or some other shape? The paraxial approximation tells you which one to use.

Once all that is set up you have to work out the geometry. In particular, for a ray that starts at the tip of the object, and ends at the tip of the image, where does it intersect the radius of the lens? Once you get all that working you can also figure out how the shape you've chosen fixes the focal length.

There are other ways to get to the answer. Some of them are codified in certain mathematical machinery, but they all are mathematical grease to apply the condition that all the optical path lengths are the same.

update after comment

A couple of questions came up in the comments.

(1) I would first invent a lens profile. I'd start with one in which the thickness decreases quadratically from the center. The optical path length from obj tip to img tip is easily calculated. Then I'd pick some other point on the radius of the lens, and calculate the optical path from obj tip to selected point to image tip, and see if the optical path length is the same as that for the "straight through" ray. I haven't tried this; if I can find time I will.

(2) Fermat's principle might be stated "The shortest distance between two points is a straight line" where "distance" means "optical path length". Consider the classic Snell's Law set up. The optical path length in the dense medium is $n$ times the geometric distance. Imagine compressing the medium by a factor of $n$ to account for this. In this new "universe" the ray is straight. It doesn't bend.

BTW, since you are digging into this, you might look into Hamilton's Point Characteristic. It's a very elegant approach to these kinds of question.

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  • $\begingroup$ Thanks a lot for the answer. Yes, I am aware of the derivation by using Fermat's principle, as you pointed out. However, I have two concerns. (1) What I learned from standard textbook, this derivation only deals with light rays emitted from the optical axis (and shows that it gets focused onto another point on the optical axis). What happens to a point shifted away from the axis (2) If one wants to start from Snell's law, is it possible to arrive the same conclusion. $\endgroup$
    – gamebm
    Jun 6, 2016 at 12:39
  • $\begingroup$ Fermat's principle is general, and will apply to rays starting anywhere. Since Snell's law follows from Fermat's principle, you should be able to use Snell's law. Note that if you use the profile from a real lens, the rays won't meet perfectly at the image because of aberrations. $\endgroup$
    – garyp
    Jun 6, 2016 at 13:50
  • $\begingroup$ My thanks and two questions. (1) How to show explicitly that a ray starting anywhere will get focus to a point (and the aberration is negligible for real lens) by using Fermat's principle. (2) Snell's law can be derived from classical electrodynamics, but what is the physical foundation of Fermat's principle? I read in Goldstein's classical mechanics (in the chapter of Hamilton equation) that it can be derived from the principle of least action, but I feel less convincing since I do not really know the Lagrange of a photon in the medium (as well as why Hamilton is conserved in this case). $\endgroup$
    – gamebm
    Jun 6, 2016 at 19:59
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    $\begingroup$ I've provided a brief response as an update to my answer. If I find some time, I'll try to expand on my comments. $\endgroup$
    – garyp
    Jun 7, 2016 at 12:44
  • $\begingroup$ Concerning the first point of the updated answer, I am thinking of the following possible argument. Consider a thin spherical lens. For a point on the optical axis, the proof (calculation of optical path) can be found in a standard textbook. Now, now instead of shifting the point away from the optical axis, let us rotate the lens for a very small angle with respect to the vertex, and try to evaluate the corresponding optical path. The difference in optical path (before and after the rotation) comes from two sections in air and one section inside the lens (all due to the rotation). $\endgroup$
    – gamebm
    Jun 7, 2016 at 13:38
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For a "thin spherical" lens, the angle as a function of height is given by

$$h = \tan^{-1}\frac{h}{d}$$

Where $d$ is the distance.

For small angles, $\tan^{-1}\frac{h}{d} = \frac{h}{d} $

For simplicity, we will do the derivation for a lens that is hemispherical (one flat surface) - this just means that the curvature is all on one surface, instead of two surfaces.

Now for a point P that is off axis, we want to calculate the shape of the surface that makes every ray that passes the lens cross the same point Q - also off axis.

enter image description here

We can look at the angles relative to the surfaces of the lens:

enter image description here

If we use a small angle approximation, then $\sin\theta=\theta$ and we can write

$$\begin{align}\alpha_1 &= \alpha_3-\theta_1\\ \alpha_2 &= \theta_2\\ \theta_1 d_1 + \theta_2 d_2 &= p + q\end{align}$$

The $\alpha$ terms can be related to Snell's Law - while the $\theta$ terms will get you to the Lensmaker's Formula. I can't write the details of that right now - but I am leaving this unfinished derivation here before I lose it. If someone feels like finishing this before I can return, they are welcome...

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  • $\begingroup$ Thanks for the hint! The shape implies $\alpha_3$ is a function of $h$, namely, $\alpha_3=\alpha_3(h)$, to be determined. Snell's law gives $n(\frac{\alpha_1}{n}+\alpha_3)=\alpha_2$, and one has additionally $p=d_1\theta_1+h$, $q=d_2\theta_2-h$ and $\frac{p}{q}=\frac{d_1}{d_2}$ from geometry. Now putting everything together one should determine the shape of the lens. But I count 10 variables but only 7 equations (9 is needed). Or I feel at least I need one more equation... $\endgroup$
    – gamebm
    Jun 6, 2016 at 19:38

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