6.8 Reflectance and transmittance

Reflectance and transmittance

Lets remember what Irradiance is: We have the Poynting vector $$\vec{S}=c^2{\epsilon }_0{\vec{E}}_0\times {\vec{B}}_0{e}^{i2\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$ But the problem is that this varies rapidly in time. So we define the Irradiance $$I={⟨S⟩}_T$$ which has units Watts per meter squared and can also be called the radiant fluxdensity. We showed in lecture that this is (where I drop the T because it is tiresome to write) $$<S>=c^2{\epsilon }_0{E}_0{B}_0/2$$ which can also be written $$<S>=c{\epsilon }_0{E}_0^2/2$$ We need to make one minor modification though, the above presumes we are in freespace. So we modify the irradiance to take into account that different media have different speeds of light. (Now we write $\text{v}$ instead of $c$ because we are not assuming free space) $$I=<S>=\frac{\text{v}\epsilon }{2}{E}_0^2$$ The reflectance is the ratio of the reflected power to the incident power is $$R=\frac{I_{r}A\cos {\theta }_r}{I_{i}A\cos {\theta }_i}$$ $A$ is the area illuminated by the electomagnetic radiation and ${\theta }_r,{\theta }_i$ are the reflected and incident angles. But we know that ${\theta }_r={\theta }_i$ so $$R=\frac{\frac{{\text{v}}_r{\epsilon }_r}{2}{E}_{0r}^2}{\frac{{\text{v}}_i{\epsilon }_i}{2}{E}_{0i}^2}$$ and the media are the same so $$R={\left(\frac{E_{0r}}{E_{0i}}\right)}^2=r^2\text{.}$$

Likewise the transmittance (using ${\mu }_0\approx {\mu }_t\approx {\mu }_i$ ) is $${\displaystyle \begin{array}{c}T=\frac{I_{t}A\cos {\theta }_t}{I_{i}A\cos {\theta }_i}\\ =\frac{I_t\cos {\theta }_t}{I_i\cos {\theta }_i}\\ =\frac{\frac{{\text{v}}_t{\epsilon }_t}{2}{E}_{0t}^2\cos {\theta }_t}{\frac{{\text{v}}_i{\epsilon }_i}{2}{E}_{0i}^2\cos {\theta }_i}\\ =\frac{{\mu }_0{\text{v}}_t{\epsilon }_t{E}_{0t}^2\cos {\theta }_t}{{\mu }_0{\text{v}}_i{\epsilon }_i{E}_{0i}^2\cos {\theta }_i}\text{.}\end{array}}$$ Now note $${\displaystyle \begin{array}{c}{\mu }_0\epsilon \text{v}=\mu \epsilon \text{v}\\ =\mu \epsilon \frac{1}{\sqrt{\mu \epsilon }}\frac{c}{c}\\ =\frac{\sqrt{\mu \epsilon }}{\sqrt{{\mu }_0{\epsilon }_0}}\frac{1}{c}\\ =\frac{n}{c}\text{.}\end{array}}$$ So now we can write $${\displaystyle \begin{array}{c}T=\frac{{\mu }_0{\text{v}}_t{\epsilon }_t{E}_{0t}^2\cos {\theta }_t}{{\mu }_0{\text{v}}_i{\epsilon }_i{E}_{0i}^2\cos {\theta }_i}\\ =\frac{n_t{E}_{0t}^2\cos {\theta }_t}{n_i{E}_{0i}^2\cos {\theta }_i}\\ =\frac{n_t\cos {\theta }_t}{n_i\cos {\theta }_i}{\left(\frac{E_{0t}}{E_{0i}}\right)}^2\\ =\frac{n_t\cos {\theta }_t}{n_i\cos {\theta }_i}{t}^2\text{.}\end{array}}$$ This is a more complicated expression than $R$ because1)The speed of energy transmission is affected by the medium2) ${\theta }_i\ne {\theta }_t$ so the projected areas normal to the propagation direction are different.