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Opticalpath difference formula

If $A$ represents the location of a point source then all points $P$ such that $$OPL[P \Leftarrow A]=\int_A^P \nu ds = S_A$$ for fixed $S_A$ is a *surface* representing an equal phase front in physical optics associated with the emitter at $A$ and the surface that is at an optical distance $S_A$ from it. In a homogeneous medium the equal phase front relative to a a point emitter is a sphere but if the index $\nu$ is varying that is not the case. Since integration is additive one may write $OPL[Q \Leftarrow A] = OPL[Q \Leftarrow P] + OPL[P \Leftarrow A]$ for any ray connecting $Q \Leftarrow P \Leftarrow A$ and you get the geometrical optics manifestation of Huygens's principle according to which every point on the wavefront to which $P$ belongs is another source of a ray to another wavefront containing $Q$ such that they are all at equal optical distance $OPL[Q \Leftarrow P] $ away from the wavefront $OPL[P \Leftarrow A] $. It can be shown using Fermat's principle that the rays are orthogonal to the wavefronts, ie. the direction of a ray emanating from a wavefront is parallel with the gradient of that surface at that point. Stigmatic imaging corresponds to the wavefront degenerating to a single point, in other words all ray paths connecting two fixed points have the same OPL. If both the emitter and target points are surrounded locally by a homogeneous medium then the duality of rays and wavefronts can be exploited by analyzing how much distortion of the wavefront suffers relative to the ideal stigmatic spherical one. In accordance with the above definition all points on the same wavefront are at zero optical distance away from each other.

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Stigmatic imaging corresponds to the wavefront degenerating to a single point, in other words all ray paths connecting two fixed points have the same OPL. If both the emitter and target points are surrounded locally by a homogeneous medium then the duality of rays and wavefronts can be exploited by analyzing how much distortion of the wavefront suffers relative to the ideal stigmatic spherical one.

Opticalpathlengthdefinition

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Opticalpathlengthdifference

If the path is in the reverse direction then the sign can be taken to be the opposite of the former (ie., setting $ds \to -ds$) signifying that we are integrating from $B$ to $A$. If $A$ represents the location of a point source then all points $P$ such that $$OPL[P \Leftarrow A]=\int_A^P \nu ds = S_A$$ for fixed $S_A$ is a *surface* representing an equal phase front in physical optics associated with the emitter at $A$ and the surface that is at an optical distance $S_A$ from it. In a homogeneous medium the equal phase front relative to a a point emitter is a sphere but if the index $\nu$ is varying that is not the case. Since integration is additive one may write $OPL[Q \Leftarrow A] = OPL[Q \Leftarrow P] + OPL[P \Leftarrow A]$ for any ray connecting $Q \Leftarrow P \Leftarrow A$ and you get the geometrical optics manifestation of Huygens's principle according to which every point on the wavefront to which $P$ belongs is another source of a ray to another wavefront containing $Q$ such that they are all at equal optical distance $OPL[Q \Leftarrow P] $ away from the wavefront $OPL[P \Leftarrow A] $. It can be shown using Fermat's principle that the rays are orthogonal to the wavefronts, ie. the direction of a ray emanating from a wavefront is parallel with the gradient of that surface at that point. Stigmatic imaging corresponds to the wavefront degenerating to a single point, in other words all ray paths connecting two fixed points have the same OPL. If both the emitter and target points are surrounded locally by a homogeneous medium then the duality of rays and wavefronts can be exploited by analyzing how much distortion of the wavefront suffers relative to the ideal stigmatic spherical one. In accordance with the above definition all points on the same wavefront are at zero optical distance away from each other.

I've encountered an exercise where L(A,B) = sum of other OPLs that are parts of itself, and at the end we got that L(A,B) = 0?!

Opticalpathlength

For example, for a plane mirror, with objet $A$ and image $A'$ we have : $L=(AI)+(IA')=AI - IA' = 0$ : the optical path is indeed zero in this case.

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Opticalpathlengthequation

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Opticalpath difference

In accordance with the above definition all points on the same wavefront are at zero optical distance away from each other.

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Since integration is additive one may write $OPL[Q \Leftarrow A] = OPL[Q \Leftarrow P] + OPL[P \Leftarrow A]$ for any ray connecting $Q \Leftarrow P \Leftarrow A$ and you get the geometrical optics manifestation of Huygens's principle according to which every point on the wavefront to which $P$ belongs is another source of a ray to another wavefront containing $Q$ such that they are all at equal optical distance $OPL[Q \Leftarrow P] $ away from the wavefront $OPL[P \Leftarrow A] $. It can be shown using Fermat's principle that the rays are orthogonal to the wavefronts, ie. the direction of a ray emanating from a wavefront is parallel with the gradient of that surface at that point.

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Opticalpathlengthderivation

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Opticalpath definition

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The optical path between two points $A$ and $B$ is related to the phase difference of the wave between these two points : $L_{AB}=\frac{\omega}{c}(\varphi(B)-\varphi(A))$

corresponds to the distance L(A,B) in vacuum equivalent to the distance AB traversed in the medium of index n at the same time t.

With this relation, the optical path appears as an algebraic quantity which is positive if one goes in the direction of the light, negative in the opposite direction. For the virtual extension of a ray, we consider that the light goes in the same direction as on the real part.

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Denoting by $ds$ the infinitesimal length of a ray path the optical path length between two points , say, $A$ and $B$, is the line integral defined by $$OPL[B \Leftarrow A]=\int_A^B \nu ds$$ where the integration path is along a $ray$ connecting the two points from $A$ to $B$. Note that the path is not just an arbitrary curve but rather the actual path that the ray takes in the medium.

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