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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=2c/

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)

genius_generous answered 2 years ago
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