Appendix I to Ch. 3 The Eikonal Equation in Optics

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Muriel Hawkins
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1 Appendix I to Ch. 3 The Eikonal Equation in Optics I. The Eikonal Equation of a Ray We may apply the Fermat's principle to find the path of propagation of a light ray. The optical path length propagated by a ray from point A to point B is given by L A B nds. (1) The ray would follow the path for which L op is an extremum; i.e. δ L op = δ A B nds=0. () Assume n is continuous everywhere, we may define a point characteristic function L (eikonal) given by the eikonal equation L L = n and L = n = neˆ t. (3) L op

2 Since for any closed curve Γ containing points A and B we have We find that for any path from point A to point B, we have Path 1 Path Γ L dr =0. (4) L dr = L dr. (5) A B A B If we trace a ray in the direction of L on any point of its optic path from A to B, we have L dr = L ds A B A B Ray, eˆ t Ray, eˆ t dr = nds= n dr. (6) ds A B A B Ray, eˆ t Ray, eˆ t

3 We can prove that this path will give a minimum value of L op since nds = L dr = L dr A B A B A B Ray, eˆ t Ray, eˆ t Other Rays = cos δ L ds L ds= nds. (7) A B A B A B Other Rays Other Rays Other Rays Eq. (7) tells us that if the ray propagates along a path in the direction of L (i.e., e ), its optic path-length will be a minimum ˆt in accordance with the Fermat s principle. We therefore show that the ray path will follow Eq. (3) the eikonal equation.

4 To obtain the ray equation of optics, we differentiate the eikonal equation after the path. Thereby we apply d eˆt ds = and obtain d 1 1 ( ) ( ) ( ) 1 L = eˆ t L = L L = L = n. (8) ds n n n Thus we obtain the ray equation d d dr n = L = n ds ds ds. [ Ray Equation ] (9) The physical ray path will be described byr from solving Eq. (9).

5 Example: Fata Morgana (Mirage) As a short example we will treat reflection at a hot film of air near the ground, which induces a decrease in air density and thereby a reduction of the refractive index. We may assume in good approximation that for calm air the index of refraction increases with distance y from the bottom (see Fig. E1), e.g. n(y)=n 0 (1 e αy ). Since the effect is small, ε 1is valid in general, while the scale length α is of the order α = 1 m 1. We apply the ray equation for all individual components of We find r = ( x, y( x)). d dx dx n = 0 n = C. [ x component] ds ds ds

6 Similarly, we see the solution for the y coordinate with the aid of above result and find that n( y) d dy d dy dx dx d dy C = n = n = C y ds ds dx dx ds ds dx dx n d dy n( y) 1 dn C C = n = dx dx y dy We may rescale the x coordinate to adjust the constant C as 1 without loss of generality. Since ε 1, we get 0 0 ( α y ) ( α y ε ε ) n = n 1 e n 1 e.. Thus dn dy n e. y 0ε α

7 Fig. E1 We find d y y = n 0εe α. dx This equation can be solved by fundamental methods and it is convenient to write the solution in the form 1 { ( ) } 0 ln cosh α y y = y + κ x x0 n0ε e. α κ = ε αn0e α y 0 /

8 For large distances from the point of reflection at x=x 0 straight propagation as expected. The maximum angle φ =tan 1 (κ/α) is defined by we find κ = ε αn e α y 0 / 0. As in Fig. E.1 the observer registers two images one of them is upside down and corresponds to a mirror image. The curvature of the light rays declines quickly with increasing distance from the bottom and therefore may be neglected for the 'upper' line of sight. At (x 0, y 0 ) a 'virtual' point of reflection may be defined.

9 II. The Eikonal Equation as a Consequence of Fermat Principle We can obtain the ray equation from the Fermat principle by applying the standard variational principle for light. The Fermat's principle, states that light will take the path which extremizes the optical path length from point A to B. The optical path length is given by L op A B n ( r ( s )) ds, (10), where I have indicated the dependence on arc length s. Following the usual variational approach, we try to find which path extremizes L op, i.e. δ L op = δ A B nds=0. (11)

10 We consider varying the path of the light ray with regard to r ( s) r ( s) + δ r ( s) with end points fixed. The first order variation in L op is given by δ L ( ) + ( ) op = δ n ds n δ ds. (1) A B It is important to note that the infinitesimal arc length changes as well. We can calculate the first δn term as δ r δ n = n dr ( ). Keep only first order terms in, the variation of arc length δds is given by ( ) ( ) dr dδ r δ ds = dr + dδ r dr = ds. ds ds

11 We thus obtain the variation of L op as dr dδ r δ Lop = n dr + n ds ds ds A B d dr = n n δ rds (1) ds ds A B The last equality there follows from an integration by part and the fact that δ r vanishes at the end points. Impose δl op =0 for arbitrary variations of δ r, we achieve the extremal path. This implies that the extremal path satisfies d dr n n = ds ds 0. [Ray Equati on ] (13)

12 We propose to define the point characteristic function L (eikonal) as the physical optical path for a ray to transverse from a point P 0 to any point P, that is L( P) L( P0 ) = n( r ) ds. (14) Since dr n( r ) ds= n( r ) dr, ds P P P P 0 0 P 0 P We arrive at L dr = L ( P ) L ( P ) P P 0 dr = n( r ) ds= n( r ) dr, (15) ds P P P P 0 0 which yields the relation dr ( ) L = n( r ) and L = n [ Eikonal Equation] (16) ds 0

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