# Reflectance or reflectivity spectrum interpretation

Hello forum,

The reflectance spectrum has % on the vertical axis and wavelength on the horizontal axis.

A red object is red because it reflects all the red light and absorbs all the other wavelengths of the white incident light.

If the read object was ideal, would its reflectance spectrum be a delta at a specific wavelength and reflectance value 100%?

I guess the reflectance spectrum is obtained by scanning the reflected energy at each wavelength for incident white light.

Does the area under the reflectance function have value unity?

Some fluorescent substances have reflectance spectrum with values above 100%. What happens at those wavelengths?

Thanks,
Fog37

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Simon Bridge
Homework Helper
Check the description that comes next to the graph - I would expect the graph to show the proportion of incident light at that wavelength that is reflected.
Thus, with florescent materials, you get >100% for some wavelengths because more light of those wavelengths is reflected than is incident, so it glows with that color rather than just being that color.

You should be able to work out for yourself whether the area of the graphs you have add up to 100% (not unity because you have a percentage on the vertical...)

Thanks Simon Bridge.

So, let's say that 100 J of incident light (distributed over the visible spectrum, as it happens for sunlight) falls on the object. If wavelength_1 has reflectivity 80%, it means of that energy is reflected at that wavelength. If another wavelength has 30% it means that 30% of the incident is reflected. But that cannot be since wavelength one leaves only 20% of incident energy to be either absorbed or reflected by other wavelengths but surely not 30%...

Do you see my confusion? Unless we measuring the spectrum this way: we shine 100 J and use a narrowband filter to only look at the reflectivity due to lambda_1. We repeat the same procedure at all other wavelengths.

Thanks,
Fog37

I imagine what you're looking at is essentially like a reflection efficiency. In your example, wavelength_1 will reflect 80 J of light at that particular wavelength.

At wavelength_2 it will reflect 30 J. The remaining 70 J will be split between being absorbed or transmitted through the material.

This is because materials have different emissivities at different wavelengths. For example, typical window glass is clear in the optical wavelengths (400-700nm? I forget), while it is opaque in the far IR wavelengths (at least 8+ um).

The total area under the curve gives you an idea how effective of a reflector an object is across all wavelengths.

Fluorescent objects work by taking wavelength at L_1 and reemitting it at L_2 (where L_1 < L_2). Basically you get a photon that comes in, interacts with an electron and kicks it into a higher energy level. It'll sit up there for some amount of time, and eventually fall back down into a lower energy state. When this happens, a lower energy photon is emitted, and those photons can be of a visible wavelength. What generally happens is UV light in -> Visible Out. The wavelength of light which is reemitted is very well defined (due to the QM nature of orbital energies), so you tend to get very vivid colors. The fact it can take seconds to minutes for the electron to decay in orbit is why you can have pigments that quite literally, glow in the dark.

thanks RacinReaver.

Let me see if I really understand it correctly: the reflectance spectrum and its %percent values should be interpret at each wavelength.

Case 1: starting with 100 J of incident monochromatic light all contained at lambda_1. Assume the reflectance spectrum has two peaks only, one 80% peak at lambda_1 and one peak of 30% at lambda_2. what happens? 80 J will be reflected (the other 20 J will be absorbed and converted into heat). No reflected light at lambda_2.

Case 2: 100 J of incident light (40 J contained in lambda_1 and 60 J in lambda_2). What happens? 32 J (which is 80% of 40J) will be reflected with lambda_1 and 18J will be reflected at lambda_2. The remaining energy corresponding to (100-32-18)= 50 J will be converted completely into energy.

What do you think? i think I get it now.

Why does the reflectance spectrum depend on the direction from which the observer is looking at the object?
Why does the reflectance spectrum depend on the surface topography of the object and not only on the chemical composition of the object?

thanks,
fog37

Light coming in has three choices.

1) Reflect
2) Transmit through object
3) Be absorbed

These are represented by a number between 0 and 1.

It's somewhat difficult to think of how a material can do all three, but you can do a fairly simple thought experiment. Think of when you shine a laser pointer right up against your finger. Looking at the incident side, you can see a lot of light is reflected. Looking on the other side, you can see some light manages to be transmitted. Looking at it end-on shows how it gets dimmer as you go further through the thickness, which represents absorption.

Now for your other two questions. I'm going to give thought experiments instead of straight out explaining them. :)

Why does the reflectance spectrum depend on the direction from which the observer is looking at the object?
Think of when you're looking at a large body of water. If you're looking at a shallow angle. For example, at the ocean from the shore, or a petri dish with a dime at the bottom. What do you see? Now look at it straight down. What do you see?

Why does the reflectance spectrum depend on the surface topography of the object and not only on the chemical composition of the object?
For this, think about what surface texture does to incoming light. Take a shiny piece of silverware, then scuff it with steel wool (don't actually do this). What did you do to the surface, what sort of reflection do you get now? Why is it different?

As a final note, remember that each of these properties (emissivity, transmissivity, and reflectivity) are dependent upon both wavelength and temperature. A great example I've had to be concerned about was a germanium window. In the visible spectrum, Ge is opaque. However, in the 8-14 um wavelength (far IR), it is highly transparent (a 3mm thick sheet is something like 94% transparent). However, if the window gets too hot, the transmissivity drops to nearly zero, and instead it absorbs all the radiation. What does this do? Instead of letting the IR through, it'll absorb it. This raises the temperature of the window, making it even better at absorbing IR. Soon enough you'll have runaway heating and a broken window, leading to a catastrophic leak in your vacuum system. :)