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My daughter and I found a box of old transparencies (clear plastic sheets) and something struck me as odd: even though each transparency is clear, the whole stack of transparencies looks whitish - why is that?

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Note that this is the same phenomenon that happens when you look at a roll of clear packing tape, which I have already thought through.

When you hold a single sheet up, it’s clear; I expect a small reflectance because only a small amount of light is being reflected from the front and back surfaces, while most of it is being transmitted through. However, holding the whole stack will decrease the transmittance because there are more surfaces to reflect light back into the incident medium. However, your question is why the whitish appearance?

When I think of white it means that all of the frequencies of light are being reflected. That is, the many surfaces of the transparencies are acting mirror-like. Since each transparency surface is smooth, you are getting specular reflections (all visible frequencies are reflected). In addition, the index of refraction for plastic does not vary much over the visible frequency range, therefore, the plastic reflects all visible light almost identically (I am essentially quoting David here). Since you are getting the entire incident frequencies reflected back into the incident medium, the transparencies will appear white.

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  • $\begingroup$ I think it is necessary to mention that it appears white and not shiny, because the plastic surface is too irregular to create a mirror-like shining effect. Instead it disperses the light. $\endgroup$
    – udiboy1209
    Aug 12 '13 at 15:29
  • $\begingroup$ @udiboy: I disagree with you. It is shiny just like “shiny metal” - try it. Although implicit in the explanation, the many interfaces produce a larger number of closely spaced specular reflections acting as they were frequency-independent. This is exactly what happens with a metal surface being shiny – this frequency-independent reflectance. $\endgroup$
    – Carlos
    Aug 12 '13 at 16:09
  • $\begingroup$ Thanks Carlos. Your explanation was easy to understand and the stack of transparencies is shiny. $\endgroup$
    – Helen
    Aug 12 '13 at 16:37

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