How much amount of light a shining surface absorbs
Answers
Answer:
I have a very basic doubt: If we shine light on a metal surface, it is well known that frequency components exceeding the threshold frequency of the metal get absorbed to contribute to photo-electrons. What happens to the frequencies lower than the threshold? Are they completely reflected? or a part of it is absorbed till the skin depth which leads to heating up of the metal? In the latter case, is there any way to calculate how much fraction of radiation is reflected and how much is absorbed (excluding PE effect)?
Answer:
Outside that range most of the incident EM field is reflected.
Because in metals n^2+k^2>>1, the Fresnel equations for normal incidence may be expressed simply as:
R=[(n-1)^2+k^2]/[(n+1)^2+k^2]
Observe that for small n (real component of the refractive index), such as the case is for metals in the visible range, R is close to unity.
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Hope it helps :)