# Solution to 5.2-3: Attenuation of Light

Ex | = | exp – | w · k · x c | · exp[ i · (kx · x – w · t)]
--- | --- | --- | --- | ---
|  | Decreasing amplitude | Plane wave
|  |  | Starting from the equation at right Give maximal values for k (damping constant, attenuation index, extinction coefficient) if a penetration depth of 1m, 100 m 104 m is specified for the light intensity. |

| The intensity I is proportional to E2 and thus decreases with I = I0exp – {(2w · k · x)/c}.
|  | We have ln{I/I0} = –(2w · k · x)/c, or k = –{c/2x w}ln{I/I0}. If we assume that I/I0 = 1/e as a measure of still sufficient intensity, we have ln (I/I0) = –1 and thus obtain k = (1/x) · (c/2w).
|  | Taking w = 1016 s–1 we have c/2w = 3 · 108 ms–1 / 2 · 1016 s–1 = 1,5 · 10–8 m. We arrive at the following table for k = 1,5 · 10–8/x for x given in meter (m).
x | k
--- | ---
1 m | 1,5 · 10–8
100 m | 1,5 · 10–10
104 m | 1,5 · 10–12
| Obviously the imaginary part of the complex index of refraction needs to be rather small if we talk about "optics". If it would be 1,5, i.e. about the same as the real part, the penetration depth would be 10–8 m = 10 nm—we have a rather opaque material in this case.
| Calculate what that would mean in terms of only e'' or only e''.
| From the definition of the complex index of refraction we have 2k2 = (e' 2 + e'' 2)½ – e' .
|  | For small e'' we can develop the square root into a series and get (e' 2 + e''2)½ » e' + e''/2e' .
|  | This leads to e'' = 4k2 · e'. For a reasonable e' = 2 (giving n = 1,4) we have e'' = 1,8 · 10–15; 1,8 · 10–29, 1,8 · 10–23 for the three k values from above.
| To relate k to the static conductivity sstatic we use the equation sDK/ 2e0w = n k, which gives
|  | sDK = 2n k e0w = k · 2,8 ·1016 · 8,854 · 10–12 s–1A·s·V–1m–1 = k · 2,48 · 105 (Wm)–1
|  | Plugging in the numbers for k we get
k | sDK [Wm–1] | rDK = 1/sDK [W cm]
--- | --- | ---
1,5 · 10–8 | 3.72 · 10–3 | 2,69 · 104
1,5 · 10–10 | 3.72 · 10–5 | 2,69 · 106
1,5 · 10–12 | 3.72 · 10–7 | 2,69 · 108
|  | For large penetration depth we need pretty good DC insulators. For 100 km = 105 m, something a fibre optic cable should do, we have about 3 GWcm, a number that is not too large for good insulators like glass. Of course, the static resistivity is not the only reason for absorption.

| Discuss the results with respect to the complex index of refractions of Si and the dielectric function of GaAs as given in this link for frequencies above and below the band gap (after you located the band gap by straight thinking)
| First we look at the GaAS curves with a linear scale.
|  | For energies below the bandgap, GaAs is transparent and both k and e'' should be zero. That's the case for energies < 1.4 eV as easily seen in the linear GaAs diagrams with the eV scale. The correct value is 1,42 eV, so optical measurements do provide easy access to bandgaps.
| Or do they?
|  | The log scale picture for Si is better suited to look for precise numbers. First, we need to convert wavelengths to eV via E = hn = hc/l. A big drop in k for Si occurs around 400 nm , corresponding to (4,1356 · 10–15 eVs · 3 · 108 ms–1)/400 · 10–9 m = 3,1 eV ????
|  | Aha! We are far off the proper value of 1,1 eV. Obviously the proper wavelength must be about twice as large, 847 nm to be exact. That is still in the infrared as is should be since Si wafers are not transparent to any visible wavelength.
|  | What we learn is that the bandgap does show up very clearly in optical measurement (look at the linear scales!) but that precision measurements might be more tricky than naively expected.

With frame

Complex Index of Refraction of Silicon

Exercise 5.2.3 Attenuation of Light

Exercise 5.2.1 Fresnel Coefficients

Exercise 5.2.4 Fresnel Equations and Polarization

Exercise 5.2.5 Rayleigh Scattering

© H. Föll (Advanced Materials B, part 1 - script)