Titanium dioxide has high refractive index and it is one of the whitest materials.

  • Does refractive index help in giving $\rm TiO_2$ white color? I know it appears white because it scatters all the visible light incident upon it.

  • Does this scattering have any thing to do with high refractive index?


I would like to add a point to Stephan's answer:

White titanium dioxide is made up wavelength-scale particles, which scatter the light in a relatively wavelength-independent (Mie-type) manner. This scattering does rely on the refractive index being larger than one. Case in point, if the particles had $n=1$, they would be invisible (in air). This also relies on the TiO$_2$ refractive index having a very small imaginary part (absorption would make the film black or otherwise colored).

So while the large index of titanium dioxide does not make it "whiter" than other non-absorbing nanoparticles (in the sense of having a flatter scattering spectrum), you likely need a thinner layer of TiO$_2$ compared to other, lower index, materials to get the same scattering amplitude (opacity).

Update regarding comment:

There was a question regarding Mie scattering for dielectric nanoparticles. It turns out that Mie theory is often the framework of choice to describe light scattering from wavelength-scale particles, and it makes the physicist's favorite approximation: treating the particles like spheres. These particles can have any permittivity (i.e. dielectric, conducting, etc) as long as they have ~spherical symmetry, since the theory is really more about the spherical expansion of scattered waves than anything else.

As an example, I found a non-paywall paper in which the authors use Mie theory to compute scattering coefficients for TiO$_2$ nanoparticles with the aim of designing white surfaces.

  • $\begingroup$ Wow :) I'm astonished to find Mie scattering in here. +1 , if you add a reference for that? I'm just familiar with metallic nanoparticles with Mie scattering. $\endgroup$ – Stefan Bischof Aug 28 '17 at 13:49
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    $\begingroup$ No problem! Answer edited. If you're interested, there are plenty of other examples of Mie scattering used to describe dielectric material systems. $\endgroup$ – Gilbert Aug 29 '17 at 1:37

Refractive index ist connected to reflection by Fresnel equations

  • Reflection (specular) describes "light bouncing" in a specific direction. Commonly simplified as "incoming angle equals reflecting angle".
  • diffuse reflection (see BRDF and its connection to surface roughness, Mie scattering) is dominiating in titandioxide $TiO_2$. You should study it more closely.

Refractive index is a different thing commonly discussed in context of lenses. Answer:

  • Yes. specular reflection is connected by Fresnel equations. s-polarized reflection is given by $$R_s=\left|\frac{n_1\cos\vartheta_i-n_2\cos\vartheta_t}{n_1\cos\vartheta_i+n_2\cos\vartheta_t}\right|^2$$
  • No, diffuse reflection and refractive index are not (strictly) connected. This is the dominating answer for titandioxide $TiO_2$.

Roughness <span class=$R_a$ and BRDF angles">

Depicted is the surface roughness average $R_a$, which is mostly used. Do not confuse surface normal $n$ with refractive index $n_1$ or $n_2$.

An interesting advanced thought, why I wrote "not strictly": High refractive index points to low transmission inside the material, hence less subsurface scattering. This might assume a minor role for titandioxide $TiO_2$.

  • $\begingroup$ I would assume that the Fresnel equations can be used as an ingredient to calculate the BRDF as long as the surface is locally smooth on the length scale of the wavelength of the light. But I have to add that I don't know how the surface of titanium white looks like. Nevertheless I think that the BRDF is kind of a phenomenological tool that tells you about the reflectance properties of the material. It does not tell you in detail why the material has these properties. For this you have to take a closer look and I would assume that the refractive index then shows up again. $\endgroup$ – Gregor Michalicek Aug 27 '17 at 17:28
  • $\begingroup$ BRDF is based on scattering(Huygens principle), not in reflection. $\endgroup$ – Stefan Bischof Aug 27 '17 at 17:34
  • $\begingroup$ You are missing some O's in your "titandioxide"s... $\endgroup$ – Steeven Aug 29 '17 at 12:35
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    $\begingroup$ @Steeven I catch some O's and bind them. $\endgroup$ – Stefan Bischof Aug 29 '17 at 12:55

I think there are many contributing aspects to the whiteness of titanium white. I think it is difficult to cover all of them in a single answer. You already have a very nice answer from Stefan Bischof who emphasizes the importance of the BRDF in this case.

I want to mention another aspect. The refraction index is obtained as

$n = \sqrt{\mu_r\epsilon_r}$

where $\mu_r$ is the relative magnetic permeability and $\epsilon_r$ is the relative electric permitivity or dielectric constant. These quantities are in fact not so constant but functions of the frequency $\omega$ of the electromagnetic wave. Let us assume that the relative magnetic permeability is 1. This should be realistic for nonmagnetic materials. $\epsilon_r$ also is not a real function, but a complex function $\epsilon_r(\omega) = \epsilon_r^{'}(\omega) + i \epsilon_r^{''}(\omega)$. As a consequence also the refraction index is a complex function.

The important thing in this description is that the absorption of the electromagnetic wave in the material is determined by the imaginary part of the dielectric function. For the case of TiO2 over the whole frequency range of the visible light this is obviously rather small.


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