Question

Light of intensity 2 Wm^–2 is traveling through air (n = 1.0) is normally incident on a dielectric surface. If the intensity of the reflected light is 0.3 W m^–2 what is the refractive index of the dielectric material?

Answer 1

The reflectivity R represents the fraction of the incident light that is reflected at the interface
R=I/I₀,
where I₀ and I are the incident and reflected beams intensities respectively.
If the light is normal (or perpendicular) to the interface, then
R={(n₂-n₁)/(n₂+n₁)}²,
where n₁ and n₂ are the indices of refraction of the two media.
When light is transmitted from a vacuum or air into a solids, then
R={(n₂-1)/(n₂+1)}²,
Therefore,
I/I₀ = {(n₂-1)/(n₂+1)}²,
0.3/2 =0.15 ={(n₂-1)/(n₂+1)}²,
(n₂-1)/(n₂+1) = sqrt0.15 =0.39
n₂-1 = 0.39(n₂+1)
0.61 n₂ = 1.39
n₂= 2.28