- #1
indie452
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okay so the question involves a boundary between 2 dielectrics.
one with n1=1.5 the other n2=1.4
incident angle of wave I= 80deg
critical angle i found to be C= 69deg
given wavelength in air is 700nm
1) prove component of wave vector parallel to boundary is = ky = k2*(n1/n2)*sin(I)
...this i have done
2)Prove that if the wave incident angle I>C then kx in the second dielectric is purely imaginary...this i have done
3)If I=80deg calculate the depth in the second dielectric at which the amplitude of the electric field has fallen by a factor of exp(-5) from its surface value
...THIS I HAVE NOT DONE
i was told in lectures that if a wave has an absorption coeff = (ni*w)/c then the, the absorption length is = c/(ni*w) but i don't know what to do from here
one with n1=1.5 the other n2=1.4
incident angle of wave I= 80deg
critical angle i found to be C= 69deg
given wavelength in air is 700nm
1) prove component of wave vector parallel to boundary is = ky = k2*(n1/n2)*sin(I)
...this i have done
2)Prove that if the wave incident angle I>C then kx in the second dielectric is purely imaginary...this i have done
3)If I=80deg calculate the depth in the second dielectric at which the amplitude of the electric field has fallen by a factor of exp(-5) from its surface value
...THIS I HAVE NOT DONE
i was told in lectures that if a wave has an absorption coeff = (ni*w)/c then the, the absorption length is = c/(ni*w) but i don't know what to do from here