Hello everybody!
I read that if one of the two materials involved in Snell's law is a conductor, the refraction angle [itex]\theta_r[/itex] is about [itex]\pi / 2[/itex] and is independent of the incident angle [itex]\theta_i[/itex] (I think [itex]\theta_r[/itex] will be precisely [itex]\pi / 2[/itex] if the conductor is ideal). My question is: why?
[itex]\theta_i[/itex] and [itex]\theta_r[/itex] are real quantities. The only other variables are the refractive indexes. How is the refractive index in a conductor? I found that we could write [itex]\epsilon[/itex] from the IV Maxwell's equation such that it includes the conduction:
[itex]\nabla \times \mathbf{H} = j \omega \mathbf{D} + \mathbf{J} = j \omega \epsilon \mathbf{E} + \sigma \mathbf{E} = [/itex]
[itex] = j \omega \epsilon ' \mathbf{E} + (\omega \epsilon '' + \sigma) \mathbf{E} = [/itex]
[itex] = j \omega \left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right) \mathbf{E}[/itex]
So [itex]\left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right)[/itex] is a new complex dielectric constant; its imaginary part takes into account the conduction. We can still write
[itex]\epsilon = \epsilon_0 \epsilon_r[/itex]
by considering
[itex]\epsilon_r = \displaystyle \frac{\epsilon '}{\epsilon_0} - j \frac{\epsilon ''}{\epsilon_0} - j \displaystyle \frac{\sigma}{\omega \epsilon_0}[/itex]
Now the refractive index is [itex]n_2 = \sqrt{\epsilon_r}[/itex] and is complex. But how it must be to generate in the Snell's law a [itex]\theta_r = \pi / 2[/itex], despite of [itex]\theta_i[/itex]?
[itex]\displaystyle \frac{n_1}{n_2} = \frac{\sin \theta_r}{\sin \theta_i}[/itex]
[itex]n_1[/itex] is the refractive index of a dielectric. If [itex]| \epsilon '' + \sigma / \omega | \gg \epsilon '[/itex] (this happens in a good conductor), we will have [itex]n_2 \gg n_1[/itex] and
[itex]\sin \theta_r \simeq 0[/itex]
while I expected [itex]\sin \theta_r \simeq 1[/itex]. What's wrong?
Thank you anyway!!
Emily
I read that if one of the two materials involved in Snell's law is a conductor, the refraction angle [itex]\theta_r[/itex] is about [itex]\pi / 2[/itex] and is independent of the incident angle [itex]\theta_i[/itex] (I think [itex]\theta_r[/itex] will be precisely [itex]\pi / 2[/itex] if the conductor is ideal). My question is: why?
[itex]\theta_i[/itex] and [itex]\theta_r[/itex] are real quantities. The only other variables are the refractive indexes. How is the refractive index in a conductor? I found that we could write [itex]\epsilon[/itex] from the IV Maxwell's equation such that it includes the conduction:
[itex]\nabla \times \mathbf{H} = j \omega \mathbf{D} + \mathbf{J} = j \omega \epsilon \mathbf{E} + \sigma \mathbf{E} = [/itex]
[itex] = j \omega \epsilon ' \mathbf{E} + (\omega \epsilon '' + \sigma) \mathbf{E} = [/itex]
[itex] = j \omega \left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right) \mathbf{E}[/itex]
So [itex]\left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right)[/itex] is a new complex dielectric constant; its imaginary part takes into account the conduction. We can still write
[itex]\epsilon = \epsilon_0 \epsilon_r[/itex]
by considering
[itex]\epsilon_r = \displaystyle \frac{\epsilon '}{\epsilon_0} - j \frac{\epsilon ''}{\epsilon_0} - j \displaystyle \frac{\sigma}{\omega \epsilon_0}[/itex]
Now the refractive index is [itex]n_2 = \sqrt{\epsilon_r}[/itex] and is complex. But how it must be to generate in the Snell's law a [itex]\theta_r = \pi / 2[/itex], despite of [itex]\theta_i[/itex]?
[itex]\displaystyle \frac{n_1}{n_2} = \frac{\sin \theta_r}{\sin \theta_i}[/itex]
[itex]n_1[/itex] is the refractive index of a dielectric. If [itex]| \epsilon '' + \sigma / \omega | \gg \epsilon '[/itex] (this happens in a good conductor), we will have [itex]n_2 \gg n_1[/itex] and
[itex]\sin \theta_r \simeq 0[/itex]
while I expected [itex]\sin \theta_r \simeq 1[/itex]. What's wrong?
Thank you anyway!!
Emily