Evaluate reflectance of an aluminum mirror

AI Thread Summary
The discussion focuses on calculating the reflectance of an aluminum mirror using the Drude model. The user calculates the angular frequency of light at 500 nm and finds it to be less than the plasma frequency, suggesting full metallic reflection (R=1). However, the expected reflectance is stated as 99%, prompting questions about potential errors in the calculation. The conversation highlights the limitations of the Drude model, which may not account for losses, and suggests that additional parameters like conductivity and atom density could be relevant. The user plans to consult their tutor for clarification on the appropriate model to use.
mcas
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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 &lt; \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 &gt; \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?
 
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The symbol ω is used to denote angular frequency in radians/second. The frequency f is related to ω by ω=2πf. So the angular frequency of light of wavelength λ is given by:
\omega = \frac{2 \pi c}{\lambda}
 
mcas said:
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?
If you are meant to be using the Drude (free electron) model, then you are correct.

The frequency of 500nm light is less than aluminium’s plasma frequency (even allowing for possible confusion/error due to the use of the symbol ‘##\omega##’ for a value specified in Hz). So the model predicts R=1.

All I can think of is that the Drude model has (significant) weaknesses. Better models will predict that R is slightly less than 1 due to losses. Perhaps that’s why you are given the conductivity and atom-density - maybe you need to use these values.

However, I’m not familiar with the models so I'm sorry that I can’t help more than that.
 
Steve4Physics said:
If you are meant to be using the Drude (free electron) model, then you are correct.
The Drude model was the only one we've discussed during the lecture, so I would expect it is used in the solution. Thank you for clarifying this! I'll need to ask my tutor whether we should use something different that wasn't covered yet.
 
If you correct your 2π error, do you get the right answer?
 
phyzguy said:
If you correct your 2π error, do you get the right answer?
\omega= \frac{2\pi c}{\lambda}=\frac{2 \cdot \pi \cdot3\cdot 10^8}{500 \cdot 10^{-9}}\approx 3.77\cdot 10^{15} Hz

Which is still less than \omega_p so R=1.
 
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