Tuesday, 7 April 2015

optics - Why does a glass window reflect white objects white from an atomic scatterers’ viewpoint?


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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