Distance travelled by wave at whichamplitude decreases

[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 don't know what to do from here

 

[indie452]

ok so I've tried this:

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

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

therefore absorption coeff = 5k_i
absorption length = 1/5k_i = c/5n_iw

k_i = n_iw/c
w = 2pi*c/n₂L...L = wavelength = 700nm

so absorption length = n₂L/n_i10pi = 3.1194x10^-8/n_i

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