present“ p” case and with the above procedure( summarized by Eqs.( 6)–( 9)) we are able to specify them.
Indeed, for completeness, we provide here the explicit mathematical expressions for the components of the electric field of the three waves, assuming the incident amplitude to be 1 for simplicity. See Appendix for more details. Incident field:
E~ i ðÞ¼1 ~ r exp i ~ k i ~ r
8 >< 1 cos h i
¼ 0 >:
�1 sin h i Reflected field:
Transmitted field:
exp i ~ k i ~ r
; ~ � k i ¼ k x i; 0; kz i: ð10Þ
See Equation( 11) bottom of this page
See Equation( 12) bottom of this page Note that, according to( A. 15), k z t in equation( 12) may be complex-valued. 2.3 Checking the formulas
For this purpose, we compare the results in equations( 6)–( 9) with those obtained using equations( 14) and( 15) in [ 10 ]. For instance, the first term in( 8)
t jj
X
cos h t
¼ t jj ¼ cos h i
X
E jj t
X
E jj i
¼
2n i cos h t n i cos h t þ n t cos h i
;
2n i cos h i n i cos h t þ n t cos h i cos h t cos h i
ð13Þ
corresponds exactly in [ 10 ] to( despite slight differences in notation)
t x p Ex t
E x i
¼ e ik z t e t k z t y p ¼ e ik z t
2e t k z i e t k z i i e t k z i þ e ik z ¼ 2e ik z t t e t k z i þ e ik z t
8 >< 1 cos h i r jj
E~ r ðÞ¼ ~ r 0 exp
>:
1 sin h i r jj
J. Eur. Opt. Society-Rapid Publ. 21, 21( 2025) 207
: ð14Þ
i ~ k r ~ r
The equality t jj
X
¼ t x p
, follows directly from the basic identities( see Appendix for the details):
e i ¼ n 2 i; e t ¼ n 2 t; kz i ¼ k 0 n i cos h i k z t
¼ k 0 n t cos h t: ð15Þ
Similarly, the following equivalences can also easily be shown: t jj
Z
¼ t z p; r jj X
¼ r x p; r jj Z
¼ r z p: ð16Þ
In summary, the direct procedure using the complex angle approach( Eqs.( 6)–( 9)) is equivalent to the more formal developments presented in [ 10 ], with the advantage of being more straightforward. Of course, the methods outlined in [ 9 ] can also be applied. However, we believe that our approach is faster and, hopefully, more intuitive.
We will now demonstrate how equations( 6)–( 9) are employed to solve and interpret the general case of an isotropic plane interface, with particular focus on incidence beyond the critical angle and on configurations involving an absorbing second medium. In both cases, additional information about the transmitted wave is required for completeness( see Appendix and [ 10 ]).
3 Two examples
We plan to graphically illustrate the behavior of the fields at the interface in two non-trivial cases.
3.1 Internal reflection
Consider the interface between two transparent media with refractive indices n i = 1.5, n t = 1.0, with the incidence angle h i = 55 °. Since the incidence is above the critical angle, we have total reflection. The ratios between tangential and normal components of the electric field are given by the formulas( 7)–( 9). Here they give, in detail,
r jj ¼ 1:0 exp 2pi
56:34 360
See Equation( 18) bottom of this page
�; ~ kr ¼ k x r; 0; kz r; with k x r kx i; kz r �kz i:
; ð17Þ ð11Þ
8 ><
E~ t ðÞ¼ ~ r >:
1 cos h i t jj
X
0 �1 sin h i t jj
Z
exp i ~ k t ~ r
�; ~ kt ¼ k x t; 0; kz t; with k x t
k x i: ð12Þ
8 ><
>:
t jj
cos h t
�61:83
X
¼ t jj ¼ 1:42 exp 2pi cos h i
t jj
Z
¼ t jj sin h t sin h i
¼ 1:42 exp 2pi
0: 71 exp½2pið90 = 360ÞŠ
¼ 1:76 exp 2pi 360
0:573
�61:83 1: 23 360 0:82 ¼ 2: 12 exp 2pi �61:83
360
28:17 360
:
; ð18Þ