The following set up with a source, 2 convex lenses, a slit and a screen is of that of Fraunhofer's diffraction:

enter image description here

*Correction S is not on the common optical axis, but above it.

$\theta$ = Angle between ray passing through midpoint(B) and the common optical axis(which also passes through B)

$a$ = Length of AC (A and C being the extremes of the slit)

When finding the point of 1st minima, we equate the optical path difference between 2 rays passing parallel through the slit at a distance of $\frac{a}{2}$ to $(2n-1)\frac{\lambda}{2}$, i.e $\frac{a}{2} \sin\theta = \frac{\lambda}{2} $

But according to Fermat's Principle of Least time, don't all the paths take the same amount of time(the least time = $\tau$) to travel from S to P and hence the optical path should be the same for all the paths = $c\tau$ , so why is there a difference in optical path?

  • $\begingroup$ Fermat's principle gives the ray approximation to the optical field and this approximation ignores all wave effects such as difraction and interference. $\endgroup$
    – mike stone
    Commented Apr 5 at 18:24

1 Answer 1


If measured from the source you take the constant phase surfaces of the propagating wave, these are the wavefronts, then Fermat's principle holds between any to points that lie on the same orthogonal trajectory of these wavefronts. Here by an orthogonal trajectory, also called an optical ray, is meant the tangential envelope to the gradients of the constant phase surfaces. As an example, the constant wave surfaces emanating from a point source in a homogeneous medium are concentric spheres centered around the source and the orthogonal trajectories are the radial lines.

Fermat's principle applied to wave propagation would then be that the propagation time between two fixed points on the same orthogonal trajectory is the shortest among all paths connecting those points if the whole path is to lie on that very trajectory. If there be several such paths, hence trajectories, then they all must have the same propagation time. This is what in optics is called stigmatic imaging of a focus to a focus.

  • $\begingroup$ So the different rays (ex: passing through A,B,C and the other points in the slit) that interfere at point P are all part of different orthogonal trajectories? $\endgroup$
    – soccerer
    Commented Apr 6 at 17:53
  • $\begingroup$ that is correct. $\endgroup$
    – hyportnex
    Commented Apr 6 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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