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 =2k_i=2n_i /c ---------(2)

where = angular frequency=2c/

And Thus, n_i=imaginary part of refractive index of silver=c/2(from 2) ------(3)

Thus, n_i=c/4c=/4 -------(4)

Since,=587 nm and =3.9 (given)

n_i=(3.9 *587*10^-9)/(4*3.14)=182.177*10^-9

n_r= freal part of ref index=0.05

The phase angle for reflection is given by

=tan^-1(2nn_i/n_r²-n²+n_i²)

=0^o (since n_i is very small)

genius_generous answered 1 year ago

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