# Distance travelled by wave at whichamplitude decreases

1. Apr 18, 2010

### indie452

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 dont know what to do from here

2. Apr 18, 2010

### indie452

Re: distance travelled by wave at which amplitude decreases

ok so ive tried this:

E=Eo*exp[i(wt-kz)]
=Eo*exp[i(wt-krz)]exp[-kiz)] - shows amp falls by exp[-1]

so
=Eo*exp[i(wt-krz)]exp[-5kiz)] - shows amp falls by exp[-5]

therefore absorption coeff = 5ki
absorption length = 1/5ki = c/5niw

ki = niw/c
w = 2pi*c/n2L...........L = wavelength = 700nm

so absorption length = n2L/ni10pi = 3.1194x10-8/ni

but how do i find ni and is this right anyway?