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A light wave falls normally on the surface of glass coated with a layer of transparent substance. Neglecting secondary reflections, demonstrate that the amplitudes of light waves reflected from the two surfaces of such a laver will be equal under the condition n' = sqrt(n), where n' and n are the refractive indiced of the layer and the glass respectively. |
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Answer» Solution :Suppose the incident light can be decomposed into waves with intensity `I_(||) & I_(bot)` with oscillations of the electric vectors parallel and perpendicualr to the palne of incidence. For NORMAL incidence we have from Fresnel equations `I'_(bot) = I_(bot) ((theta_(1) - theta_(2))/(theta_(1) + theta))^(2)-rarrI_(bot) ((n - 1)/(n+1))^(2)` where we have used `theta~~ theta` for small `theta`. SImilarly `I'_(||) = I_(||) ((n' - 1)/(n'+1))^(2)` Then the refracted wave will be `I''_(||) = I_(||)(4n')/((n' + 1)^(2))` and`I''_(bot) = I_(bot)(4n')/((n' + 1)^(2))` At the interface with glass `I'''_(bot) = I''_(bot) ((n' - n)/(n' + n))^(2)`, similarly for `I'''_(||)` we see that `(I'_(bot))/(I_(bot)) = (I'''_(bot))/(I''_(bot))` if `n' = sqrt(n)`, similarly for `||` COMPONENT. This shows that the light REFLECTED as a fraction of the incident light is the same on the two SURFACES if `n' = sqrt(n)`. Note:- The statement of the problem given in the book is incorrect. Actual amplitudes are not equal, only the reflectance is equal. 
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