PerformativeDesign.com
Definitions
Light
Reflectivity and Reflectance

Reflectivity and Reflectance

Many opaque materials have the same values for reflectance and reflectivity, so people tend to use the terms interchangeably. However, there are many other materials for which the value can be very different so it is important to properly understand what reflectance and reflectivity really mean.

Because reflectivity is a boundary property and is not affected by material depth (beyond the nanometer scale), many reasonably thick opaque building materials have the same value for both reflectance and reflectivity. However, it is very important to note that these are actually very different things and the terms are not interchangeable.

Reflectivity is the property of the boundary between two particular types of media whilst reflectance is the property of the boundary of between two particular instances or samples of those media. For most opaque building materials of measurable thickness, reflectance values essentially equals reflectivity values. However, for transparent, translucent or very thin materials there are additional effects such as internal reflections from the rear face or scattering from internal particulates that can result in very different values for overall reflectance and reflectivity.

For non-opaque materials, reflectivity can be thought of as the lower limit of reflectance, or the reflectance resulting only from the very first intersection of the light with the surface. Any other internal reflections or emissions serve only to increase the overall reflectance whilst the reflectivity remains the same.

Investigating in more detail specific cases of why reflectance and reflectivity differ can help in understanding the the two terms.

Glass Example

Window glass is a key material where reflectance can differ from reflectivity. This occurs primarily because of internal reflections between the rear and front faces of the glass pane. The reflectance and reflectivity have the same value only at the very first intersection with the boundary, where some light is reflected and some transmitted. However, the transmitted portion also then hits the rear face of the glass where, again, some light is reflected and some transmitted. In this instance, the reflected portion travels back through the glass until it hits the front face where, again, some light is reflected and some transmitted. This internal reflection happens many times, as shown in Figure 3.

Figure 1: An illustration of the multiple internal reflections that occur in glasses that result in reflectance and transmittance values significantly different from reflectivity and transmissivity. For the sake of clarity, absorption was not considered.

For the purposes of conceptual clarity, a perfectly non-absorbing glass has been assumed. In a normal glass, there would be some amount of absorption of the transmitted portion, so the transmitted/reflected portions immediately beneath the upper surface would be slightly higher that the transmitted/reflected portions actually reaching the lower surface.

This example shows that, even though the reflectivity is 0.074, the total reflectance due to multiple internal reflections is 0.1378.

Similarly, because some of the transmitted energy is reflected back internally, even though the transmissivity is 0.926, the total transmittance is only 0.8622.

To fully explain this diagram, lets go through the calculations involved.

1. Calculate Angle of Refraction

The first step is to find the angle of refraction for the light or radiation transmitted into the second material using Snell’s law:

$$ n_2 \sin \theta_x = n_1 \sin \theta_i $$

This can be rearranged to give the angle of refraction:

$$ \theta_x = \sin^{-1} \left( \frac {n_1 \sin \theta_i } {n_2} \right) $$

where:

  • θi is the angle of incidence, in degrees or radians, and
  • θx is the angle of refraction, in degrees or radians.
  • n1 is the refractive index of the medium the light is travelling from,
  • n2 is the refractive index of the medium the light is travelling into,

The specification of degrees or radians means that it depends upon which maths library you are using and how it requires you to provide angle input for the sin and cos trigonometric functions. If you are using a pocket calculator, you can usually switch modes and enter either degrees or radians. If you are implementing these equations in computer code, you will very likely need to use radians.

2. Calculate Reflectivity at the Boundary

For s-polarized light, where the wave is at 90° to the plane of incidence, the reflectivity is given by:

$$ R_s = \left| \frac {n_1 \cos \theta_i - n_2 \cos \theta_x} {n_1 \cos \theta_i + n_2 \cos \theta_x} \right|^2 $$

where:

  • Rs is the reflectivity and initial reflectance for s-polarized light,

For p-polarized light, where the wave is parallel to the plane of incidence, the reflectivity is given by:

$$ R_p = \left| \frac {n_1 \cos \theta_x - n_2 \cos \theta_i} {n_1 \cos \theta_x + n_2 \cos \theta_i} \right|^2 $$

where:

  • Rp is the reflectivity and initial reflectance for p-polarized light.

For unpolarized light with an equal mixture of s- and p- components, the results of the two equations must be averaged:

$$ R = \frac {R_s + R_p} {2} $$

where:

  • R is the reflectivity and initial reflectance for non-polarized light.

3. Calculate Transmissivity at the Boundary

$$ T = 1 - R $$

where:

  • T is the transmissivity and initial transmittance for non-polarized light,

Doing the Numbers

Now lets do the numbers for this specific example. First we find the angle of refraction if the angle of incidence is 45 degrees and the refractive index of air is 1.00029 and this particular glass is 1.67:

$$ \begin{align} \theta_x &= \sin^{-1} \left( \frac {1.00029 \times \sin(45\deg)} {1.67} \right) \\ \\ &= \sin^{-1}(0.42354002524137) \\ \\ &= 25.06\deg \end{align} $$

Next we find the reflectivity for s-polarized light:

$$ \begin{align} R_s &= \left| \frac {n_1 \cos \theta_i - n_2 \cos \theta_x} {n_1 \cos \theta_i + n_2 \cos \theta_x} \right|^2 \\ \\ &= \left| \frac {(1.00029 \times \cos(45\deg)) - (1.67 \times \cos(25.06\deg))} {(1.00029 \times \cos(45\deg)) + (1.67 \times \cos(25.06\deg))} \right|^2 \\ \\ &= \left| \frac {0.70915739085199 - 1.51279409136781} {0.70915739085199 + 1.51279409136781} \right|^2 \\ \\ &= \left| \frac {-0.80363670051582} {2.2219514822198} \right|^2 \\ \\ &= \left| -0.3616805798626 \right|^2 \\ \\ &= 0.13081284184975 \end{align} $$

Next we find the reflectivity for p-polarized light:

$$ \begin{align} R_p &= \left| \frac {n_1 \cos \theta_x - n_2 \cos \theta_i} {n_1 \cos \theta_x + n_2 \cos \theta_i} \right|^2 \\ \\ &= \left| \frac {(1.00029 \times \cos(25.06\deg)) - (1.67 \times \cos(45\deg))} {(1.00029 \times \cos(25.06\deg)) + (1.67 \times \cos(45\deg))} \right|^2 \\ \\ &= \left| \frac {0.9061274261403 - 1.18086832458153} {0.9061274261403 + 1.18086832458153} \right|^2 \\ \\ &= \left| \frac {-0.27474089844123} {2.08699575072183} \right|^2 \\ \\ &= \left| 0.13164420595787 \right|^2 \\ \\ &= 0.01733019696228 \end{align} $$

Then we average the two because we are working with non-polarised light:

$$ \begin{align} R &= \frac {R_s + R_p} {2} \\ \\ &= \frac {0.13081284184975 + 0.01733019696228} {2} \\ \\ &= \frac {0.14814303881203} {2} \\ \\ &= 0.07407151940602 \end{align} $$

Next we find the transmissivity for non-polarised light.

$$ \begin{align} T &= 1 - R \\ \\ &= 1 - 0.07407151940602 \\ \\ &= 0.92592848059398 \end{align} $$

Tracking Internal Reflections

Now we need to apply these at each boundary intersection. For the sake of expediency and to clarify the maths, lets take the two calculated values as 0.074 and 0.926 instead of 0.07407151940602 and 0.92592848059398.

$$ \begin{align} T_1 &= 0.926 \\ R_1 &= 0.074 \\ \\ T_2 &= 0.926 \times 0.926 = 0.857476 \\ R_2 &= 0.074 \times 0.926 = 0.068524 \\ \\ T_3 &= 0.926 \times 0.068524 = 0.063453224 \\ R_3 &= 0.074 \times 0.068524 = 0.005070776 \\ \\ T_4 &= 0.926 \times 0.005070776 = 0.004695538576 \\ R_4 &= 0.074 \times 0.005070776 = 0.000375237424 \\ \\ T_5 &= 0.926 \times 0.000375237424 = 0.00034746985462 \\ R_5 &= 0.074 \times 0.000375237424 = 0.00002776756938 \\ \\ T_6 &= 0.926 \times 0.00002776756938 = 0.00002571276925 \\ R_6 &= 0.074 \times 0.00002776756938 = 0.00000205480013 \\ \\ T_7 &= 0.926 \times 0.00000205480013 = 0.00000190274492 \\ R_7 &= 0.074 \times 0.00000205480013 = 0.00000015205521 \\ \\ T_8 &= 0.926 \times 0.00000015205521 = 0.00000014080312 \\ R_8 &= 0.074 \times 0.00000015205521 = 0.00000001125209 \\ \\ T_9 &= 0.926 \times 0.00000001125209 = 0.00000001041944 \\ R_9 &= 0.074 \times 0.00000001125209 = 0.00000000083265 \end{align} $$

Looking at all the values on the same side of the material as the incident light, the total reflectance becomes:

$$ \begin{align} R &= R_1 + T_3 + T_5 + T_7 + T_9 \\ &= 0.13780260701898 \end{align} $$

Looking at all the values on the opposite side of the material as the incident light, the total transmittance becomes:

$$ \begin{align} T &= T_2 + T_4 + T_6 + T_8 \\ &= 0.86219739214837 \end{align} $$

Thin Film Example

Consider as another illustrative example a light-coloured opaque material with a high reflectance. In this material, reflections occur from molecules at the surface as well as up to 50nm below the surface. By the time light has penetrated up to 100nm into the material it is totally absorbed, which is why the material is opaque.

Now, let’s take a very thin film of that material, say 25nm thick. In this instance, some of the light that would have been absorbed is now able to pass through it. More importantly, some of the light that would have been reflected back out by molecules deeper than 25nm now passes through instead. Thus, the overall reflectance is now reduced as the transmittance has increased, but the reflectivity of the material stays exactly the same. This may sound like a very unlikely edge case, but silvering or electro-coating of glass with metallic films of this kind of thickness is actually quite commonplace.

Useful References

Click here to comment on this page.

© 2016 Dr. Andrew Marsh. Some rights reserved.