When we do problems with optics and refraction, we’re usually given a set of indices of refraction to work with; for example, the index of refraction in air is about 1.00, the index of refraction in water is 1.33, etc. However, the index of refraction also varies with wavelength, and these given indices are usually based on yellow light (wavelength = 589 nm). Why is it that we use yellow light as the standard? When we do problems with refraction, why do we take the index of refraction for yellow light whenever the wavelength of light we’re dealing with is unspecified?
[Physics] Why do we use the index of refraction for yellow light
conventionsopticsrefractionvisible-light
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Here is an analogy that I like to use: (even though it is not really a correct physical explanation)
Imagine that you are out riding your segway over some strange surfaces, that each have a number $n_i$ that controls the speed that a segway wheel travels over it according to the formula $v_i=v_0/n_i$. Now imagine that you cross a straight boundary between two surfaces at an angle. Because of the angle, one wheel will cross the boundary before the other. If $n_i$ is higher for the entered surface this wheel will go slower than the other until it too crosses the boundary, which will cause the segway to turn towards the normal of the boundary. Similarly, if $n_i$ is lower for the entered surface, the first wheel to enter will go faster, and the segway will turn from the normal.
If you do the calculations for the segway you will get the the same results as for the wavefront explanation (basically Snell's law), but I really like how this analogy works with your intuition.
Electromagnetic theory combined with optics gives an electromagnetic interpretation for the refractive index: $$ v = \frac{1}{\sqrt{\epsilon_0\mu_0}\sqrt{\epsilon_r\mu_r}} = \frac{c}{\sqrt{\epsilon_r\mu_r}} = \frac{c}{n} \quad\Longrightarrow\quad n^2 = \epsilon_r\mu_r $$
If we ignore magnetic properties, ie $\mu_r\approx 1$, then we have: $n^2 = \epsilon_r$. If you solve maxwell equations for an waving electromagnetic field in vacuum, you will get this result for plane waves: $\mathbf k\times\mathbf B = -\omega/c^2\epsilon_r\mathbf E$. The form of the $\mathbf{E}$ plane wave is: $$\mathbf E = \mathbf E_0 e^{i(\omega t - \mathbf k\cdot\mathbf r)}$$
But, then, when you plug an electromagnetic plane wave into the Maxwell Equations for a non-magnetic material with conductivity $\sigma$ and relative complex permittivity $ \hat{\epsilon}_r = \epsilon'+i\epsilon''$, the current is $\mathbf J_{\text{tot}} = J_{\text{displacement}} + J_{\text{conduction}} = i\omega\epsilon\mathbf E_0 + \sigma\mathbf E_0 = i\omega\epsilon_0(\epsilon' - \frac{i\sigma}{\epsilon_0 \omega})$ and the Ampere Equation reduces to: $$ -i\mathbf k\times\mathbf B_0 = \frac{i\omega}{c^2}\left(\epsilon' - \frac{i\sigma}{\epsilon_0\omega}\right)\mathbf E_0 $$
Then, in comparison with previous result, we defined a complex dielectric constant: $$ \hat\epsilon_r = \epsilon' + i\epsilon'' = \epsilon' - \frac{i\sigma}{\epsilon_0\omega} $$
And then, we define a complex refraction index: $\hat n = n +i\kappa = \sqrt{\hat\epsilon_r}$, $\frac{kc}{\omega} = \hat{n}$ with the branch chosen so that the real exponential is decaying. Assuming a plane wave propagating in the $+x$ direction, you now get a damped wave, where the complex part of the refraction index $\epsilon''$ makes the wave decrease its electric field exponentially: $$ \mathbf E(x, t) = \mathbf E_0 e^{i \left( \omega t - kx \right)} = \mathbf E_0 e^{-\frac{\omega}{c}\kappa x} e^{i\left(\omega t - \frac{\omega n}{c}x\right)} $$
Best Answer
For many materials the change in refractive index over the range of visible wavelengths isn't huge, so it's not a bad approximation to take a single value. The range of visible wavelengths is from about 400nm to 700nm, so the middle wavelength is 550nm. As it happens, the sodium D lines are not far from this, at 589nm, and since they are bright and easy to produce they've become an unofficial standard for quoting the refractive index.
As CuriousOne says in a comment, for precise work you would use a graph of refractive index against wavelength (or some empirical fit). Where we need to take some account of the refractive index change, but it doesn't have to be that accurate, a rough measure of the dispersion is given by the Abbe number. Manufacturers will often quote the refractive index at 589nm plus the Abbe number.