Question

A light falls on the sample with n₁ = 1.653 and n₂ = 2.35 × 10^-2 (n = n₁ + i n₂), with a wavelength of 500 nm. Calculate the followings:

a) the speed of light in the sample
b) Wavelength of light in the sample
c) The distance at which the wave intensity is reduced by half.
d) Reflection coefficient of light from sample

Answer 1

In real materials, the polarization does not respond instantaneously to an applied electric field which causes dielectric loss. The complex index of refraction can be defined as

n^* is the complex refractive index. n₁ is the refractive index. n₂ indicates the amount of absorption loss (often called the extinction coefficient) when the EM wave propagates through the material. In most cases, n₂> 0 (light is absorbed).

(a) The speed of light in the sample, v (the velocity of light in some medium) = velocity of light in a vacuum (c)/the index of refraction of that medium (n₁) = (3 x 10⁸ m/s)/1.653 = 1.815 x 10⁸ m/s

(b) If a EM wave travel from one medium to another medium, the frequency will remain same but wavelength changes according to index of refraction. If are the wavelength in a medium and in free space, then the refractive index is related to the wavelength as given below (f is the frequency of light)

In this problem, wavelength is 500 nm and n is 1.653

Hence, the wavelength in the medium will be

(c) Under single scattering approximation, according to Beer-Lambert law, we can calculate the light intensity I after travelling a distance z attenuated by the medium with extinction coefficient n₂ (initial light intensity I₀),

I/I₀ = exp(-zn₂)

In this problem, I/I₀ = 1/2 =0.5 and n₂ = 2.35 × 10^-2

0.5 = exp(-z x 2.35 × 10^-2)

Hence, z = 29.49

(d) Reflection coefficient (R) of light from sample can be calculated from Fresnel's Equations

considering the normal incidence of light and n₁, n₂ are the index of refraction of two medium