91Ó°ÊÓ

Write the expression for the reflection coefficient for light to conducting surface.

$\mathit{R}\mathbf{=}{\mathbf{\left|}\frac{{\stackrel{\mathbf{~}}{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{R}}}}{{\stackrel{\mathbf{~}}{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{I}}}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{=}{\mathbf{\left|}\frac{\mathbf{1}\mathbf{-}\stackrel{\mathbf{~}}{\mathbf{β}}}{\mathbf{1}\mathbf{+}\stackrel{\mathbf{~}}{\mathbf{β}}}\mathbf{\right|}}^{\mathbf{2}}$ …… (1)

Here,$\mathrm{β}$ is the complex quantity which is given as:

$\begin{array}{c}\widetilde{\mathrm{β}}=\frac{{\mathrm{μ}}_1{v}_1}{{\mathrm{μ}}_2\mathrm{Ó\neg }}{\tilde{k}}_2\\ \widetilde{\mathrm{β}}=\frac{{\mathrm{μ}}_1{v}_1}{{\mathrm{μ}}_2\mathrm{Ó\neg }}(k_2+i{K}_2)\end{array}$ …… (2)

Write the value of the wave number$k_2$ .

$k_2=\mathrm{Ó\neg }\sqrt{\frac{{\mathrm{Î\mu }}_2{\mathrm{μ}}_2}{2}}{[\sqrt{1+{\left(\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_2\mathrm{Ó\neg }}\right)}^2}+1]}^{\frac{1}{2}}$

Write the value of the wave number$k_2$ .

$K_2=\mathrm{Ó\neg }\sqrt{\frac{{\mathrm{Î\mu }}_2{\mathrm{μ}}_2}{2}}{[\sqrt{1+{\left(\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_2\mathrm{Ó\neg }}\right)}^2}−1]}^{\frac{1}{2}}$

As silver is a good conductor then$\mathrm{σ}>>{\mathrm{Î\mu }}_2\mathrm{Ó\neg }$, .

The value of$k_2$ and $k_2$will be,

$\begin{array}{c}{k}_2=\mathrm{Ó\neg }\sqrt{\frac{{\mathrm{Î\mu }}_2{\mathrm{μ}}_2}{2}}\sqrt{\left(\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_2\mathrm{Ó\neg }}\right)}\\ K_2=\mathrm{Ó\neg }\sqrt{\frac{{\mathrm{Î\mu }}_2{\mathrm{μ}}_2}{2}}\sqrt{\left(\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_2\mathrm{Ó\neg }}\right)}\end{array}$

Hence, the value of$k_2$ and $k_2$becomes,

$k_2=K_2=\sqrt{\frac{\mathrm{Ó\neg }{\mathrm{μ}}_2\mathrm{σ}}{2}}$

Substitute $k_2=K_2=\sqrt{\frac{\mathrm{Ó\neg }{\mathrm{μ}}_2\mathrm{σ}}{2}}$in equation (2).

$\begin{array}{c}\widetilde{\mathrm{β}}=\frac{{\mathrm{μ}}_1{v}_1}{{\mathrm{μ}}_2\mathrm{Ó\neg }}\sqrt{\frac{\mathrm{σ}{\mathrm{μ}}_2\mathrm{Ó\neg }}{2}}(1+i)\\ \widetilde{\mathrm{β}}={\mathrm{μ}}_1{v}_1\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_2\mathrm{Ó\neg }}}(1+i)\end{array}$ …… (3)

Here,

$\begin{array}{c}\mathrm{γ}={\mathrm{μ}}_1{v}_1\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_2\mathrm{Ó\neg }}}\\ \mathrm{γ}={\mathrm{μ}}_{0}c\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_0\mathrm{Ó\neg }}}\end{array}$

Substitute $\mathrm{γ}={\mathrm{μ}}_1{v}_1\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_2\mathrm{Ó\neg }}}$and $\mathrm{γ}={\mathrm{μ}}_{0}c\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_0\mathrm{Ó\neg }}}$in equation (3).

$\begin{array}{c}\widetilde{\mathrm{β}}={\mathrm{μ}}_{0}c\sqrt{\frac{\mathrm{σ}}{2{\mathrm{μ}}_0\mathrm{Ó\neg }}}(1+i)\\ \widetilde{\mathrm{β}}=c\sqrt{\frac{\mathrm{σ}{\mathrm{μ}}_0}{2\mathrm{Ó\neg }}}(1+i)\end{array}$

Substitute$c=3×{10}^8\text{″¾}/\text{s}$ ,$\mathrm{σ}=6×{10}^7\text{ }{\left(\mathrm{Î\copyright }\text{m}\right)}^{−1}$ , ${\mathrm{μ}}_0=4\mathrm{Ï€}×{10}^{−7}\text{ H}/\text{m}$and $\mathrm{Ó\neg }=4×{10}^{15}{\text{ s}}^{−1}$in the above expression.

$\begin{array}{c}\widetilde{\mathrm{β}}=(3×{10}^8\text{″¾/s})\sqrt{\frac{(6×{10}^7\text{ }{\left(\mathrm{Î\copyright }\text{m}\right)}^{−1})(4\mathrm{Ï€}×{10}^{−7}\text{ H/³¾})}{2(4×{10}^{15}{\text{ s}}^{−1})}}(1+i)\\ \widetilde{\mathrm{β}}=29(1+i)\end{array}$

Substitute i$\widetilde{\mathrm{β}}=29(1+i)$n equation (1).

$\begin{array}{c}R={\left|\frac{1−29(1+i)}{1+29(1+i)}\right|}^2\\ R=\frac{{(1−29)}^2+{\left(29\right)}^2}{{(1+29)}^2+{\left(29\right)}^2}\\ R=0.93\end{array}$

Therefore, the reflection coefficient for light at an air air-to-silver interface is$93\%$ .