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I am reading "Optics" by E. Hecht, and I find the explanation of Fermat's Theorem a bit confusing. Specifically:

I have this illustration attached. Why in picture (b) for group-I do we have phasors so close, and for group-II we have have phasors pointing in completely different directions?

For me both groups look pretty similar in picture (a). So why do they have different phasors?

enter image description here

Qmechanic
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Valeria
  • 143

3 Answers3

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For group I, the total path length from S to P is almost the same: where line 1 is a bit shorter from S to the reflection, it is almost exactly made up for by the fact that the segment mirror-P is longer.

On the other hand, for group II, the path length from S to mirror is almost the same for all three rays; however, there is a difference in path length from mirror to P.

Floris
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The direction of the vector is a result of the total distance. Changes in the location of the angle near the center do not contribute much to changes in the total distance.

Small changes are magnified when the location of reflection is far from the center point. Thus, an equal location change results in a larger total distance change when the location of reflection is further away from the center. This larger difference means that the resulting vectors may be in wildly different directions, depending upon the wavelength.

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First recall below a beautiful geometric proof of the law of reflection i.e. incoming angle=outgoing angle using Fermat's principle by flipping the outgoing path in the mirror plane:

$\uparrow$ Fig. 1 The law of reflection. The shortest path between $A$ and the mirror point $B'$ is a straight line. (Image from the quantapublication.wordpress.com website.)

Obviously, the shortest path is a straight line. Or equivalently, the straight line is a stationary point of path lengths in the set of all paths.

  1. On one hand, the path lengths of neighboring paths to the straight line can only vary in the second order of the deformation. Therefore the corresponding phasors all point in approximately the same direction, and their sum adds up, cf. group-I in Fig. (b).

  2. On the other hand, away from a stationary point, the path length of neighboring paths typically vary linearly in the deformation. The corresponding phasors points in very different directions, and their sum typically cancels, cf. group-II in Fig. (b).

It should be stressed that above observation is at the core of why the path integral is dominated by classical paths, cf. e.g. this Phys.SE post and links therein.

Qmechanic
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