*Consider a pn junction GaAs LED. Assume that photons are generated uniformly in all directions in a plane perpendicular to the junction.*

**Part A:** *Taking into account total internal reflection calculate the fraction of photons that have the potential of being emitted from the semiconductor.*

**Part B:** *Using the results of part (a) and including Fresnel loss, determine the fraction of generated photons that will be emitted from the semiconductor into air (neglect absorption losses).*

**Part A**For the following problem we can base on Snell's law for the critical angle, before we have total internal reflection, which is:

(4)

Thus as we have been given the refractive indexes or GaAs - 3 and for air - 1 we have:

(5)

Having the critical angle we can thus relate it to the total circle of 360 , which will give us the fraction of photons having the potential of being emitted:

(6)

However we should take the other side of the plane possibility, thus we should simply multiply the result x2. Therefore the end potential of photons being emitted is 8,809 %.

**Part B**The fraction of generated and emitted photons (escaping from the LED) will depend on the reflections inside the device, thus a very commonly used formula (not only for optics but for electronic transmission lines as well) for the reflection coefficient is defined as:

(7)

which will be for our case:

(8)

Now having the fraction of photons that have the potential of being emitted and the reflection coefficient, then the fraction of photons emitted will simply be:

(9)

A short comment on the results, it seems that probably a proper geometry and configuration of the pn junction is crucial for the delivered efficiency from the device. Getting a reflection loss of roughly 3% while the photons with potential of being emitted is 8 % speaks by itself that due to the difference in refraction indexes of the materials used, the device loses a lot of its efficiency.