Imagine three light beams are "sent" to a lens simultaneously, they start at the same position but move towards the lens at different angles. The first light beam passes the lens at its edge, the second light beam passes exactly through its mid whilst the last light beam passes somewhere in between. Behind the lens, there is a screen at the focal point of the light beams. Which of the three light beams reaches the screen first?

I assume they all reach it at the same moment, but I can't explain why, it's just my intuition. Thus, I hope somebody can help me out here. Thanks for answers in advance.

//e: Sorry for being inaccurate. This is how the setup described should be like:
This http://img6.imagebanana.com/img/ahfdqbrn/IMG_1208.jpg

  • 1
    $\begingroup$ You have not specified if the beam are focused to the same point or not, and that is critical to the answer. $\endgroup$ – dmckee --- ex-moderator kitten Mar 16 '11 at 18:00
  • $\begingroup$ It sounds to me like your setup is impossible. If they all start at the same point, and one goes directly to a lens, while another one has to go to the edge of the lens, isn't the second one travelling a further distance? And if so, isn't it going to take more time to get there? $\endgroup$ – Jerry Schirmer Mar 16 '11 at 18:01
  • $\begingroup$ It would also depend on exact geometry of the setup, as well as refractive index of the lens. You can get any result you want by varying these parameters. $\endgroup$ – TROLLHUNTER Mar 16 '11 at 18:02
  • 2
    $\begingroup$ You can argue by Fermat's principle . ;-) $\endgroup$ – Sam Mar 16 '11 at 18:04
  • $\begingroup$ @Jerry: It will indeed take longer to get to the lens, but all the rays will take the same time to get to the image point. That is, in fact, the exact function of the lens. My answer expands on that a little. $\endgroup$ – Colin K Mar 16 '11 at 18:13

If the lens focuses perfectly, so that all rays from the given starting point end up at the same final point, then you're right: they all arrive at the same time. (To be precise, this is true within the limits of applicability of geometrical optics. But if we're not willing to assume that geometric optics applies, then it doesn't make sense to talk about individual rays of light anyway.)

The reason is Fermat's principle, otherwise known as the principle of least time. In a situation like this, each ray follows a path of least time joining the given starting and finishing points. If the times weren't the same, then at least one of them wouldn't be a minimum.

You might worry that there's a swindle here. Fermat's principle really only says that the path has to be a critical point, not that it has to be a minimum. But that doesn't affect the conclusion. Take two of the rays, and imagine a family of rays smoothly joining one to the other. Fermat's principle says that the derivative of the travel time is zero all along the family of rays, so there's no difference between the time at the beginning and the time at the end.

That last paragraph is a bit awkwardly phrased. Let me restate it precisely. Let $T(r)$ be the travel time corresponding to ray $r$. Let $r(0)$ be one of the rays under consideration, $r(1)$ be another, and $r(s)$ for $0<s<1$ be a smooth function joining the two together. Each $r(s)$ is a ray striking the lens at a different point, but all obeying the correct laws of optics. Fermat's principle says that $\delta T=0$ is constant for all small variations about one of these rays, so $dT(r(s))/ds=0$, and $T(r(0))=T(r(1))$.

  • $\begingroup$ All rays indeed end up in the very same final point. I apologize for not mentioning that in my question. $\endgroup$ – Huy Mar 16 '11 at 18:33

EDIT I wrote this under the assumtion that the screen was placed at the plane where the rays come to focus. I now realize that the question does not state that. If the screen is not at the focal point, then no, the rays will not have equal travel time. END EDIT

Your intuition is correct, each ray would reach the screen at the same time. In fact, in some sense, you could define the image plane of any imaging system as the plane where all rays from the object plane have an equal travel time. That would be a bit of a sloppy definition and it isn't quite complete in the technical sense, but for an intuitive explanation it will work in most cases.

Speaking in terms of geometrical optics, optical engineers usually refer to the optical path length (OPL) rather than the travel time of the light, but they are equivalent quantities, differing only by a factor of the speed of light (c). The OPL in a medium of refractive index $n$ is defined as:

$$ \text{OPL}=nL$$

Where $L$ is the actual distance traveled in the medium. So, for a ray passing through several diferent materials (like some air, a lens, and then air again) the total OPL can be expressed:

$$ \text{OPL}=\sum_in_iL_i$$

This quantity should be equal for each ray when it is properly focused from the object point to the image point.

This makes intuitive sense when you think about it. A lens with positive power is thicker at the center than the edges, and typical glass has $n\approx1.5$. The actual length of the ray through the center of the lens is shorter than the rays that go through the edges of the lens, because it follows a more direct path; but the central ray also goes through the thicker part of the lens, which increases its OPL to match that of the rays through the edge.

That pretty much sums up the geometrical optics picture.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.