Question

b) Show that the transmitted intensity, IT, though a thin transparent material of thickness l, is given by ?? =?0(1−?)2?−??, where, I0 is the incident intensity, R is the reflectance, and β is the absorption coefficient. If the material is polycrystalline describe how the transmission of light is altered by the material.

Answer 1

Let R be the reflection coefficient of the material. Then, the transmitted electric field amplitude will be:

E_t = E_o[1 - R]

since Intensity (time-averaged Poynting flux magnitude) is dependent of square of Electric field, therefore squaring both sides will give,

I_t = I_o[1 - R]²

now, this transmitted light attenuates exponentially inside the material whose attenuation is governed by Beer's law

therefore, final transmitted intensity at a distance l is given by:

where is the attenuation coefficient of the material.

Polycrystalline materials have high reflectance and so the transmitted intensity will be low.