A planar LED fabricated from GaAs has a refractive index of 2.5. Compute the optical power emitted when transmission factor is 0.68.
A)3.4 %
B)1.23 %
C)2.72 %
D)3.62 %

Question

Quizplus · Accepted Answer

The optical power emitted, often referred to as the extraction efficiency, can be calculated using the formula for the transmission factor at the interface between two media, which is given by $T = \frac{4n}{(n+1)^2}$, where $n$ is the refractive index of the material. Rearranging this formula to solve for $T = 0.68$, we get $0.68 = \frac{4 \times 2.5}{(2.5+1)^2}$, which simplifies to $0.68 = \frac{10}{3.5^2}$, or $0.68 = \frac{10}{12.25}$, which does not directly give us the percentage. To find the correct answer, we use the given transmission factor directly as it represents the fraction of optical power emitted. Since the transmission factor is 0.68, and we want to express this as a percentage, we multiply by 100 to get $0.68 \times 100 = 68\%$. However, the correct calculation method involves understanding that the transmission factor given does not directly translate to the optical power emitted without considering the critical angle and total internal reflection phenomena in planar LEDs. The correct approach involves more complex calculations involving the refractive index and the geometry of the LED. Given the options and the misunderstanding in the calculation, the correct answer based on the provided options and assuming a direct correlation would be to find the closest match to the theoretical calculation of emission efficiency based on the refractive index, which is not straightforwardly provided in the question. The correct answer choice, therefore, is based on typical extraction efficiencies for LEDs with high refractive indices without direct calculation from the given transmission factor.

A Planar LED Fabricated from GaAs Has a Refractive Index