# Evaluate reflectance of an aluminum mirror

• mcas
However, the answer says that the correct value is 99%. Could somebody explain where have I made a mistake?f

#### mcas

Homework Statement
Evaluate reflectance of an aluminum mirror for a wavelength $\lambda=500 nm$. Assume that Al conductivity is $\sigma=4.1\cdot 10^7 \Omega^{-1} m^{-1}$. The concentration for atoms per unit volume is equal to $6\cdot 10^{28} m^{-3}$ and plasma frequency is $\omega_p = 2.1\cdot 10^{16} Hz$.
Relevant Equations
$\omega_p = \sqrt{\frac{Ne^2}{\epsilon_0 m*}}$
$R=1$ 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 $\omega > \omega_p$
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?

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
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.

• mcas
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?

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 $R=1$.