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Name: Ricky L.
Status: student
Age: 18
Location: N/A
Country: N/A
Date: 3/1/2004


Question:
Dear sir, I have a problem to relate the optical density with the frequency of the visible light. I cannot explain why violet colour light has a higher refractive index than a red colour light by using the idea of optical density. Could you explain it to me?


Replies:
Hello Ricky.

Your problem is not clear to me either. "Optical density" usually means absorption amount or rate, like "how dark are your sunglasses?", not the index of refraction. (Although, if they chucked all the old terms and re-invented them all, I bet that index would be called optical density.)

So I wonder if your question has something transparent but colored in it, like clear red plastic or blue sky? This something would have absorption amounts, in percent lost total, or percent lost per meter, or similar units, for each wavelength. This optical density would have different values at different wavelengths according to some formula you have been given in class or in the question. Or maybe they are just looking for "more" of red/blue and "less" of blue/red.

Optical density can also refer to optical losses by scattering by fog or similar. In that case the "lost" light is never absorbed and destroyed, just deflected out of the straight line it was traveling as it entered the foggy or milky substance. Once the light is pointed somewhere else, it is often considered useless. For example, it no longer carries any picture information it started with. The scattering which causes the blueness of the sky is like this, and it has a wavelength dependence too. Hopefully your class told you something about that, to help you finish the problem.

Jim Swenson



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