I am solving the following problem:
Consider a GaAs emitter. Calculate the transmittance for the interface GaAs/glass/air. Assume near
normal incidence. GaAs: n= 3.4; glass: n=1.5. Answer: 81.6%
My attempt:
I know that the reflectance and transmittance are given by:
\begin{align}
R&=\left\lvert\frac{n_1-n_2}{n_1+n_2}\right\rvert\\
T&=1-R
\end{align}
Then I calculated for both interfaces GaAs/glass and glass/air
\begin{align}
R_1&=0.387755102\\
T_1&=0.612244898\\
R_2&=0.2\\
T_2&=0.8
\end{align}
I don't know to relate both transmittances to obtain the global one.
Best Answer
You sure you use the correct equations? Because for normal incidence, it is: $$R={\left\lvert\frac{n_1-n_2}{n_1+n_2}\right\rvert\\}^2$$
$$T=1-R$$
Reference - Fresnel Equation for Reflectance
Based on above equations, for GaAs-Glass interface, you will get - $$R_1=0.15$$ $$T_1=0.85$$
For Glass-Air interface, you will get - $$R_2=0.04$$ $$T_2=0.96$$
So mathematically, the effective transmittance should be -
$$T_{eff}=T_1 \times T_2 = 0.816 = 81.6 \% $$
therefore effective reflectance should be - $$R_{eff}=1-T_{eff}=0.184=18.4\%$$
If you plug in the relationship between \$R\$ and \$T\$, you can derive relation for effective reflectance as well - $$R_{eff}=1-T_{eff}$$ $$R_{eff}=1-(T_1\times T_2)$$ $$R_{eff}=1-((1-R_1)\times (1-R_2))$$ $$\implies R_{eff}=R_1+R_2-R1.R2$$