Page 157 - Photonics Essentials an introduction with experiments

Lasers

Lasers 151

By letting x become small, we can write:
d
I(x) = – I(x)
dx
I(x) = I 0 e – x W-cm –2 (7.6)
The value of depends on the material and on the photon energy. For
4
example, in the case of GaAs, is about 10 cm –1 for photons having
an energy greater than the band gap (E g = 1.43 eV at room tempera-
ture). For photons that are less energetic than the band gap energy,
is three orders of magnitude smaller. When the absorption coefficient
is a positive number, the intensity of the beam decreases as the light
propagates through the material. However, suppose that could be
made to be negative, what would be the result?

Excercise 7.1
A beam of light of monochromatic photons of energy 1.5 eV strikes the
surface of GaAs at normal incidence. What percentage of the original
photon beam penetrates 1 µm beneath the surface? What percentage
penetrates 10 µm beneath the surface?
Solution
There are only two things that can happen to photons incident on an
interface. Either they are reflected or transmitted. Some of the trans-
mitted photons are subsequently absorbed. To answer these questions
you need to find first of all the percentage of light that is transmitted,
and then find out what fraction of those photons are absorbed.
The percentage of light reflected is calculated from Fresnel’s equa-
tion at normal incidence:
(n – 1) 2 (2.4) 2
R = = = 0.25
(n + 1) 2 (4.4) 2
T = 1 – R = 0.75. So, 75% of the light is transmitted, and I 0 = 0.75 × in-
cident intensity. To calculate the intensity:
I = I 0 e – x
I = I 0 e –(10 4 cm –1 )x

I = I 0 e –1 for 1 m penetration, so
I = I 0 (0.37) = 0.75 · 0.37 = 0.28 × incident intensity
For 10 m penetration,
I = 0.75 · 4.6 · 10 –5 = 3.4 · 10 –5 × incident intensity

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