Question #113432

For normal incidennce at the air dielectric interface with dielectri constant Er=4. Find the fraction of energy reflected into the air?

Expert's answer

R=(n1n2n1+n2)2R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2

n1=1,n2=Er=2n_1=1, n_2=\sqrt{E_r}=2

R=(121+2)2=19R=\left(\frac{1-2}{1+2}\right)^2=\frac{1}{9}


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