# Solution to Exercise 5.2-1: Fresnel Coefficients

| Consider an unpolarized light beam hitting the surface of an isotropic glass with n = 1,5 at right angles (a = 0o).
| Derive the simplified Fresnel equations as given in the script for a = 0o
| Setting the cosines to 1 in the TE or TM case gives
TE case
Eref | = | Ein · | sinb – sina sinb + sina
--- | --- | --- | ---
| TM case
Eref | = | Ein · | sinb – sina sinb + sina
--- | --- | --- | ---
Division by sinb; sina/sinb = n
TE and TM
Eref | = – | Ein · | n – 1 n + 1
--- | --- | --- | ---
q.e.d
|  | There is thus no difference between the TE and TM case.

| How much of the light will be reflected?
| We can use the simple equations for perpendicular incidence from above, re-written for intensities:
Iref Iin |  | = | æ ç è | n – 1 n + 1 | ö ÷ ø | 2
|  | The intensity of the reflected light is thus
Iref | = Iin | æ ç è | n – 1 n + 1 | ö ÷ ø | 2
| = Iin | æ ç è | 0,5 2,5 | ö ÷ ø | 2
| = 0,04 Iin
--- | ---
|  | The reflected intensity is thus 4 % of the incoming intensity.
| What is the phase relation between incoming, reflected and transmitted light?
| To answer that question we must look at the field strength. There is a minus sign and the the phase of the reflected beam thus is phase-shifted by 180 o = p
| How does the beam leave the crystal (Intensity and polarization)?
| For the transmitted beam we have the simple relation Itr = Iin – Ire. The intensity of transmitted beam thus is 96 % of the incoming intensity for an optical material with n = 1,5.
|  | At the "exit" from the n = 1,5 optically dense material to the n = 1 less dense material a and b interchange their role or sinb/sina = n = 1,5 now. Taking this into account we get the same equations for the perpendicular incidence as above but without the "–" (minus) sign for the field strengths. This means that 96 % of 96 % (= 92,16 %) of the incoming beam exits the optical material.
|  | How about phases? Looking at the original Fresnel equations for the electrical field strength we see that in both the TE and TM case there is no sign and therefore phase jump for the transmitted wave for all angles (since a, b <= 90 o in all cases). For the reflected wave at the optically less dense medium (the wave reflected back into the interior of the optical material) there isn't a phase change either because the minus sign is no longer there.
| Now consider these questions for some polarization of the incoming light.
| No difference since we have the same equations for both the TE and TM case.

With frame

Exercise 5.1.1 Derivation of Snellius Law

Exercise 5.2.1 Fresnel Coefficients

© H. Föll (Advanced Materials B, part 1 - script)