Question 14.9: Calculate the critical angle at a semiconductor–air interfac...

For GaAs, \bar{n}_{2}=3.8 at a wavelength of \lambda=0.70 \mu \mathrm{m} and for air, \bar{n}_{1}=1.0. The critical angle is

\theta_{c}=\sin ^{-1}\left(\frac{\bar{n}_{1}}{\bar{n}_{2}}\right)=\sin ^{-1}\left(\frac{1.0}{3.8}\right)=15.3^{\circ}

Comment

Any photon that is incident at an angle greater than 15.3^{\circ} will be reflected back into the semiconductor.