# Evaluate reflectance of an aluminum mirror

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

mcas
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 $R=1$.

## 1. How is the reflectance of an aluminum mirror measured?

The reflectance of an aluminum mirror is typically measured using a spectrophotometer, which measures the amount of light reflected off the mirror at different wavelengths.

## 2. What factors affect the reflectance of an aluminum mirror?

The main factors that affect the reflectance of an aluminum mirror include the quality and smoothness of the aluminum coating, the angle of incidence of the light, and the wavelength of the light being reflected.

## 3. What is the typical reflectance of an aluminum mirror?

The typical reflectance of an aluminum mirror can vary depending on the factors mentioned above, but it is generally around 85-90% for visible light.

## 4. How does the reflectance of an aluminum mirror compare to other types of mirrors?

Aluminum mirrors have a higher reflectance compared to other types of mirrors, such as silver or gold mirrors. This is because aluminum has a high reflectivity for a wide range of wavelengths, making it a popular choice for mirrors in various applications.

## 5. Why is it important to evaluate the reflectance of an aluminum mirror?

Evaluating the reflectance of an aluminum mirror is important because it allows us to determine the quality and efficiency of the mirror. This information is crucial for ensuring that the mirror is suitable for its intended use, whether it be in scientific experiments, optical devices, or everyday objects like mirrors.

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