# Solution to Exercise 5.2-2: Fresnel Equations and LEDs

| Consider a simple light emitting diode schematically working as shown. All light is generated in a small volume as indicates an we assume that the semiconductor is fully transparent (which is not really true). The index of refraction of semiconductors is rather large; you may take it to be n=3
|  |  | The simple question is: How much (in %) of the light generated is transmitted through the front (upper) surface?

| First, it is important to make a drawing of the situation with respect to reflected and transmitted light:
|  | Light hitting the surface with an angle larger than acrit will suffer total reflection and remains inside the semiconductor where it is eventually absorbed. Light emitted "downwards might get out of the semiconductor but is absorbed in the housing.
|  | It follows that only light paths' inside the cone with opening angle 2acrit will get out to where we need them. Steric analysis gives the surface of the spherical cap belonging to the 2acrit cone to SSC=2p r2 · (1 – cosacrit).
|  | The percentage of the light intensity Iout coming out relative to the light intensity Itotal emitted into the full sphere with surface 4p r2 is thus
Iout Itotal | = | 2p r2 · [1 – cos(acrit)] 4p r2 | = | 1 – cos(acrit) 2
--- | --- | --- | --- | ---
| Now we need to determine acrit. Looking a the picture (and reversing the arrows) we write Snellius' law for total reflection as sinb=n · sinacrit=1. It follows that acrit=n–1 · arcsinb.
|  | For n=3 this gives us 19,47 o. Insertion gives Iout/Itotal=0,0286
|  | Only 2.86 % of the radiation produced is useful!!!!! We have a severe problem here!

| Suggest measures to improve that percentage.
| The first thing to do is to add a reflector at the bottom. That doubles the efficiency. Second thing to do is to put a material with an n between that of air and the semiconductor on top. This increases acrit and the beneficial effect is clear from the picture . Third thing to do is to shape you semiconductor in such a way that reflected light will get out after a second internal reflection. An inverted pyramid is a good shape for this (can you see why?). Fourth..... You get, maybe, the idea, that LEDs with an overall efficiency (electrical energy in / light energy out) of 50 % are not made "just so" but contain a lot of engineering. |  |

With frame

Exercise 5.1.1 Derivation of Snellius Law

Exercise 5.2.1 Fresnel Coefficients

Exercise 5.2.2 Fresnel Equations and LEDs

Exercise 5.2.4 Fresnel Equations and Polarization

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