From my understanding, refractive index is an indication as to how the speed of light changes when it passes through a material. However a professor has asked me how would I determine whether or not an object is transparent or not, simply by being given the refractive index (n) of light before and entering the object.

Is it as simple as saying that if n=0 after passing through the object, then it is not transparent? Therefore are most objects transparent to some degree?


4 Answers 4


Given just the real part of the refractive index of the material you can't find out if the material is transparent or not. You can only calculate how much light is reflected and how much light is refracted (i.e transmitted).

For example for light moving from air to water, let's assume the light has an angle of incidence normal to the water. Water has a refractive index of $1.333$ and the refractive index of air is about $1$.

With the Fresnel Equations you can calculate how much of the light is reflected:

$R=\left( \frac{1.333-1}{1.333+1} \right)^2 = 0.0203731 \sim 2\%$

So the transmitted part is:

$T = 1- R = 1-0.0203731 = 0.9796269 \sim 98\%$

However here you only see how much of the light is reflected. It doesn't give you any information about how much of the light is attenuated in the medium.

What you need is the complex refractive index. When light passes through a medium a part of it will always be attenuated. Thus one defines the complex refractive index as $\underline{n} = n + i\kappa.$ The imaginary part $\kappa$ is the extinction coefficient which indicates the amount of attenuation when the light passes through the medium. You can find the refractive indices and extinction coefficients of various materials on refractiveindex.info.

  • $\begingroup$ Thank you for your reply, it's helped my understanding. I believe I misunderstood my professor's question now; I think the second refractive index that he mentioned was not the refractive index of the second medium,but rather the imaginary part to the complex refractive index. Your answer has been very insightful though, so thanks again. $\endgroup$
    – S Hipson
    Commented May 11, 2017 at 19:28
  • $\begingroup$ @macco With $T$ you mean the refracted light correct (transmitted + absorbed)? If I understand it correctly even if $T$ is high compared to $R$ then the object may be opaque if a significant amount of $T$ is attenuated? $\endgroup$
    – Anton
    Commented Apr 30, 2022 at 19:23

The definition of refractive index is $n=\frac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ is the phase velocity of light in the new medium. If the medium is vacuum ($v=c$), then $n=1$.

As $v$ is the phase velocity (not the group velocity), then you could have $v>c$ without violating special relativity, and as a consequence, refractive indexes can be below of $1$, but only with this we cannot relate it with the transparency of the medium.

You have to use transmitance ($T$) and reflectance ($R$). Quickly said, those quantities are the porcentage of light that is transmitted and reflected, respectively, so $R+T=1$.

You can read here that, for normal incidence, the reflectance can be written as $R=\left( \frac{n-1}{n+1} \right)^2$ if the first medium has $n=1$ (vacuum or air). This is the graph for $R$ (y-axis) as a function of $n$ (x-axis).

You can observe that reflectivity changes with $n$, and that if your refractive index is very small (if $n \approx 0$) or very big (if $n \rightarrow +\infty$), then reflectivity is $R \approx 1$ so you have an almost opaque medium.

On the other hand, you will have an almost perfect transparent medium when $R$ is minimum, which happens when $n \approx 1$.

Edit: my answer works only for a medium with real refraction index (usually non-conductive medium). If the medium has a complex refractive index you would have to take it into account as in macco's answer. Usually complex refractive index describe a conductor (metallic medium) which is not normally transparent, although that is the general case.


This thread from 2013 doesn't exactly answer your question, but it is relevant and I think you will find it helpful:

Why are so many different types of objects white, yet appear gray when they are wet?

Specifically the answer given by Carlos on Sep 16 '13 at 12:40 regarding the appearance of salt grains.


Firstly, the refractive index is not the only factor that sets transparency or otherwise. Witness the examples of sheet ice and snow: a material's internal structure is also a strong factor.

So, it's clear we should concentrate on materials that are either crystalline or otherwise optically homogeneous.

Given you're a third year student, I think your professor may be angling for the Kramers-Kronig relationships. These are integral (Hilbert-transform) relationships between the real (absorption per unit propagation distance) and imaginary (phase per unit propagation distance, i.e. refractive index) parts of the complex propagation constant for a plane wave in a material. If $n(\omega)$ is the refractive index as a function of frequency $\omega$ and $\alpha(\omega)$ the absorption, then:

$$\alpha(\omega) = \frac{1}{\pi}-\!\!\!\!\!\!\int \frac{n(\omega)}{\omega^\prime -\omega}\,\mathrm{d}\omega^\prime$$ $$n(\omega) = -\frac{1}{\pi}-\!\!\!\!\!\!\int \frac{\alpha(\omega)}{\omega^\prime -\omega}\,\mathrm{d}\omega^\prime$$

These integral transform are also known as the Hilbert transform and its inverse. The slashed integral stands for "Cauchy Principal Value".

These relationships arise owing to stability considerations (see footnote) and mean that the refractive index as a function of wavelength/frequency is not independent of the absorption as a function of frequency. So, losses described by the Fresnel equations aside, there is no relationship between refractive index and transparency at a single wavelength, but there is an integral transform relationship between the two as functions of frequency - to within a constant. The Kramers-Kronig relationships show that, given the refractive index or absorption as a function of wavelength for all wavelengths, one can work out the other to within a constant.

Notice that they mix the refractive indices over the whole frequency range to compute the value of $\alpha$. So, for their accurate application, the KK relationships require that we know $n(\omega)$ over the whole range of frequencies where they are nonzero.

Below is shown a typical result of the above equations. The green curve shows a triangular absorption as a function of frequency, with frequency along the horizontal axis (actually, it is gain in a Raman amplifier, but the same principle applies). The phase of the amplifier, proportional to an effective refractive index for our purposes, is the blue curve. I emphasize yet again that the KK relationships only determine absorption / refractive index to within an arbitrary additive constant, so that's why the blue curve has both positive and negative values - it can be added to any constant refractive index and still be valid. This arbitrary constant is also why a single refractive index says nothing about absorption.

Kramers Kronig transformation


The KK relationships derive from stability considerations. Consider the Laplace transfer function $H(s) = H(i\,\omega)$ relating the output to input time variation of a plane wave. In a passive medium, it cannot have poles in the right half plane - otherwise these would correspond to boundlessly exponentially growing waves in the output. Nor can it have zeros anywhere: no finite thickness slab of material cannot absorb electromagnetic radiation at any complex frequency perfectly (although practically it can do a pretty good job).

Then $\alpha(s)+i\,n(s)\propto \log H(s)$ must be analytic in the right half plane. By the transformation $s\mapsto i\,s$, rotate the complex plane through a quarter turn so that frequencies $\omega$ are along the real axis, and $\alpha(\omega)+i\,n(\omega)$ must be holomorphic in the upper half plane and moreover are bounded as $\omega\to\infty$.

Under these conditions, it follows immediately that the real part must define the imaginary part, modulo a constant, and contrariwise. For suppose that there were two analytic functions $\alpha_1(\omega)+i\,n(\omega)$ and $\alpha_2(\omega) + i\,n(\omega)$ with the same imaginary part on the real axis, which are holomorphic in the closed lower half plane and which are also bounded as $\omega\to\infty$. The difference is then real on the real axis. Then, by the Schwarz Reflexion Theorem](https://en.wikipedia.org/wiki/Schwarz_reflection_principle) and uniqueness of analytic continuation, the difference must be holomorphic in the closed upper half plane too, but the difference is also bounded, hence, by Liouville's Theorem, i.e. that every bounded entire function is a constant, the real parts can only differ by a constant.

Here ends the elegance, but we know there must be a precise relationship, modulo a constant. Fiddly contour integral methods will get you the actual Hilbert transform formulas - the Wikipedia Kramers-Kronig Relationships page sketches the calculation. But I like the idea of proving there must be a unique relationship with clear and simple holomorphic function theory concepts before losing one's thoughts in calculations that people like me always get wrong anyway.


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