Question
In: Physics
metal of silver has Refractive index ( 0.05 ) at wavelength 587 nm. and it's attenuation...
metal of silver has Refractive index ( 0.05 ) at wavelength 587 nm. and it's attenuation coefficient are 3.9 . Find the reflectivity in case of vertical incidence and phase angle difference for the reflection .
Solutions
Expert Solution
Refractive index of Silver=0.05
Assuming surrounding medium to be vacuum, reflectivity for verical incidence is given by:-
R= (n-1)2/(n+1)2=(0.95/1.05)2=0.8186=81.86% ---------(1)
Further, the attenuation coefficient 
=2ki=2ni 
/c ---------(2)
where
= angular frequency=2
c/
And Thus, ni=imaginary part of refractive index of
silver=c/2(from
2) ------(3)
Thus, ni=c/4c=/4
-------(4)
Since,=587
nm and
=3.9 (given)
ni=(3.9 *587*10-9)/(4*3.14)=182.177*10-9
nr= freal part of ref index=0.05
The phase angle for reflection is given by
=tan-1(2nni/nr2-n2+ni2)
=0o (since ni is very small)
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