Question 9.22: Calculate the reflection coefficient for light at an air-to-...

According to Eq. 9.147, R=\left|\frac{\tilde{E}_{0_{R}}}{\tilde{E}_{0_{I}}}\right|^{2}=\left|\frac{1-\tilde{\beta}}{1+\tilde{\beta}}\right|^{2}=\left(\frac{1-\tilde{\beta}}{1+\tilde{\beta}}\right)\left(\frac{1-\tilde{\beta}^{*}}{1+\tilde{\beta}^{*}}\right) , where =\frac{\mu_{1} v_{1}}{\mu_{2} \omega}\left(k_{2}+i \kappa_{2}\right) (Eqs. 9.125 and 9.146).

\tilde{E}_{0_{R}}=\left(\frac{1-\tilde{\beta}}{1+\tilde{\beta}}\right) \tilde{E}_{0_{I}}, \quad \tilde{E}_{0_{T}}=\left(\frac{2}{1+\tilde{\beta}}\right) \tilde{E}_{0_{I}}                        (9.147)

\tilde{k}=k+i \kappa                             (9.125)

\tilde{\beta} \equiv \frac{\mu_{1} v_{1}}{\mu_{2} \omega} \tilde{k}_{2}                                (9.146)

Since silver is a good conductor (\sigma \gg \epsilon \omega) , Eq. 9.126 reduces to

k \equiv \omega \sqrt{\frac{\epsilon \mu}{2}}\left[\sqrt{1+\left(\frac{\sigma}{\epsilon \omega}\right)^{2}}+1\right]^{1 / 2}, \quad \kappa \equiv \omega \sqrt{\frac{\epsilon \mu}{2}}\left[\sqrt{1+\left(\frac{\sigma}{\epsilon \omega}\right)^{2}}-1\right]^{1 / 2}                           (9.126)

\kappa_{2} \cong k_{2} \cong \omega \sqrt{\frac{\epsilon_{2} \mu_{2}}{2}} \sqrt{\frac{\sigma}{\epsilon_{2} \omega}}=\sqrt{\frac{\sigma \omega \mu_{2}}{2}}, \text { so } \tilde{\beta}=\frac{\mu_{1} v_{1}}{\mu_{2} \omega} \sqrt{\frac{\sigma \omega \mu_{2}}{2}}(1+i)=\mu_{1} v_{1} \sqrt{\frac{\sigma}{2 \mu_{2} \omega}}(1+i) .

\text { Let } \gamma \equiv \mu_{1} v_{1} \sqrt{\frac{\sigma}{2 \mu_{2} \omega}}=\mu_{0} c \sqrt{\frac{\sigma}{2 \mu_{0} \omega}}=c \sqrt{\frac{\sigma \mu_{0}}{2 \omega}}=\left(3 \times 10^{8}\right) \sqrt{\frac{\left(6 \times 10^{7}\right)\left(4 \pi \times 10^{-7}\right)}{(2)\left(4 \times 10^{15}\right)}}=29 .

Then  R=\left(\frac{1-\gamma-i \gamma}{1+\gamma+i \gamma}\right)\left(\frac{1-\gamma+i \gamma}{1+\gamma-i \gamma}\right)=\frac{(1-\gamma)^{2}+\gamma^{2}}{(1+\gamma)^{2}+\gamma^{2}}=0.93 . Evidently 93% of the light is reflected.