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Related: Is a white object always white?

If you are standing in front of a glass window during the day, you can see your dim white t-shirt’s reflection in the window. The reflection is dim because only 4% of the light is reflected (assuming you’re perpendicular to the window) while the rest is transmitted through the glass. If a white object is placed in front of this glass pane, it’s obvious that white light is reflected as white and not colored. Why is this?

I vaguely remember that visible light that is reflected penetrates glass about λ/2, so that means that blue light penetrates the least while red light penetrates more. If I shine white light onto a glass window (assume RGB color model), I have to account for all three colored scatterers in roughly equal amounts of intensity to produce white. Here is what I think.

• Blue light penetrates shorter distances into the glass window’s atomic layers and recruits less atomic scatterers that back scatter more intensity blue light. I assume Rayleigh scattering is valid since wavelengths of light are much greater than the atomic spacing’s in glass, therefore $I ∝ 1/λ^4$.

• Green light penetrates deeper into the glass window’s atomic layers and recruits more atomic scatterers that back scatter more “less intensity” green light according to $I ∝ 1/λ^4$?. So the larger number of green atomic scatterers scattered less intense green light but there are more green atomic scatterers.

• In a similar fashion, red light would recruit an even larger number of red atomic scatterers (but less intense than green or blue), however, the increased number of red atomic scatterers compensates for a more overall intense red intensity.

In summary, there are roughly equal amounts of blue, green and red light reflected such that a glass window reflects white objects white. Here is what I am asking: is this correct way to view this? I was unable to find this anywhere online or on . I would greatly appreciate feedback, especially if this is not correct.

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Closely related: physics.stackexchange.com/questions/1957/…. Also of interest astronomical mirrors are front-silvered because rear-silvered glass mirrors do suffer from chromatic aberrations, it's just that they are too modest to notice in everyday applications. –  dmckee Jun 30 '13 at 16:16
    
@dmckee: I looked at this link carefully and it is related, however, it really explains it from a macroscopic point of view, not microscopic. I like your point about the silvered mirrors. –  Carlos Jun 30 '13 at 17:35
    
Yes. That's why I didn't close your questions as a duplicate of the earlier one. That said, if you know about a closely related questions when writing your post it is a good idea to link it in your questions and explain explicitly how your question differs. Reduces the chance of a mistaken close. –  dmckee Jun 30 '13 at 17:39
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2 Answers 2

Let me discuss just one aspect of your question. While your phrase "visible light that is reflected penetrates glass about λ/2" has some sound basis, it is rather irrelevant here, because light typically does not penetrate into the glass substrate of a mirror: it is reflected by the metal coating on the glass. Light penetrates metal to a so-called skin-depth (you may wish to google this term), which is much less than the wavelength for visible light and typical metals. So if the metal coating is at least a few skin-depths thick, light is reflected and partly absorbed before it reaches the glass substrate.

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You are correct. I've edited my question to reflect these changes. –  Carlos Jul 1 '13 at 23:40
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Household mirrors are silvered on the back, not the front. Anyway, the glass is nearly irrelevant under normal conditions; as the OP notes, reflection from the glass is very weak, if the ray is incident near the normal. –  Ben Crowell Jul 2 '13 at 2:18
    
@Ben Crowell: I agree that most "Household mirrors are silvered on the back, not the front." However, there are first-surface household mirrors as well (focus.visionresearch.com/?p=30 : "First surface mirrors are not as easy to find as common mirrors, but they are available.") I remember I had a first-surface mirror, but I remember that mostly because it degraded quickly:-) - probably, it had no protective layer:-) –  akhmeteli Jul 2 '13 at 4:09
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The amount of reflected light is predicted by Fresnel's equations. The equations are a little cumbersome, but they tell you that the fraction of light reflected depends on the index of refraction of the two materials, as well as the angle of incidence.

One thing though, is since the index of refraction changes with the wavelength of light, then the amount of reflected light also changes with the wavelength!

Realistically though, the index of refraction barely changes at all over the visible spectrum.

Index of refraction as a function of wavelength for different materials

In this image, the orange band is the visible spectrum. Those materials labeled "crown" are different types of glass. The range of the index of refraction for these glasses is only about 0.02 or 0.03 over the visible spectrum.

In other words, the glass barely notices the different colours. It reflects all visible light almost identically.

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