An \"anti-reflection\" coating must be less optically dense than the lens, for e
ID: 1385861 • Letter: A
Question
An "anti-reflection" coating must be less optically dense than the lens, for example, a coating with n = 1.30 if the lens has n = 1.70, so the reflected intensities (n1-n2)/(n1+n2) are similar ... Calculate the thinnest layer of this coating film that would cancel reflections of 590[nm] yellow sunlight. (caution: E-field flipping)
{follow-up : can you figure out what thicker layer would cancel yellow reflections?}
{follow2 : what wavelength would have its reflections enhanced in each case?}
PLESE SHOW WORK FOR SOLVIG, THANK YOU
4012 [nm]
3068 [nm]
2360 [nm]
1815 [nm]
1388 [nm]
501.5 [nm]
383.5 [nm]
295 [nm]
227 [nm]
173
4012 [nm]
3068 [nm]
2360 [nm]
1815 [nm]
1388 [nm]
501.5 [nm]
383.5 [nm]
295 [nm]
227 [nm]
173
Explanation / Answer
It is obvious that there are phase changes for the reflection of the light incident from the air (n=1) to the coating(n=1.31), and for the reflection of the light incident from the coating(n=1.31) to the glass (n=1.52). Hence these two phase change cancel one another. We only need to make the traveling of the light inside the coating to be of half the wavelength. That is: the minimum nonzero thickness of the coating is, 579 nm/(4n) = 579 nm/(4*1.31) = 110.496 nm
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