mcas
- 22
- 5
- Homework Statement
- Evaluate reflectance of an aluminum mirror for a wavelength [itex]\lambda=500 nm[/itex]. Assume that Al conductivity is [itex]\sigma=4.1\cdot 10^7 \Omega^{-1} m^{-1}[/itex]. The concentration for atoms per unit volume is equal to [itex]6\cdot 10^{28} m^{-3}[/itex] and plasma frequency is [itex]\omega_p = 2.1\cdot 10^{16} Hz[/itex].
- Relevant Equations
- [itex]\omega_p = \sqrt{\frac{Ne^2}{\epsilon_0 m*}}[/itex]
[itex]R=1[/itex] where [itex]\omega < \omega_p[/itex]
[itex]R=\left|\frac{\sqrt{1-\frac{\omega_p^2}{\omega^2}}-1}{\sqrt{1-\frac{\omega_p^2}{\omega^2}}+1}\right|[/itex] where [itex]\omega > \omega_p[/itex]
Note: for some reason frequency on this lecture is indicated by \omega.
I wanted to calculate the reflectance using one of these equations that were given to us during the lecture:
R=1 where \omega < \omega_p
R=\left|\frac{\sqrt{1-\frac{\omega_p^2}{\omega^2}}-1}{\sqrt{1-\frac{\omega_p^2}{\omega^2}}+1}\right| where \omega > \omega_p
I've calculated \omega:
\omega= \frac{c}{\lambda}=\frac{3\cdot 10^8}{500 \cdot 10^{-9}}=6\cdot 10^{14} Hz
So it's less than \omega_p and we should have a full metallic reflection. However, the answer says that the correct value is 99%. Could somebody explain where have I made a mistake?
I wanted to calculate the reflectance using one of these equations that were given to us during the lecture:
R=1 where \omega < \omega_p
R=\left|\frac{\sqrt{1-\frac{\omega_p^2}{\omega^2}}-1}{\sqrt{1-\frac{\omega_p^2}{\omega^2}}+1}\right| where \omega > \omega_p
I've calculated \omega:
\omega= \frac{c}{\lambda}=\frac{3\cdot 10^8}{500 \cdot 10^{-9}}=6\cdot 10^{14} Hz
So it's less than \omega_p and we should have a full metallic reflection. However, the answer says that the correct value is 99%. Could somebody explain where have I made a mistake?