# I How does an index of refraction affect the EM field?

1. May 11, 2017

### 2sin54

Hello. Say I have some refraction index n in a homogeneous material. Say I also have equations for the EM field (E and B vectors). Is it true to say that all that changes is the wavelength
$$\lambda \to \frac{\lambda_0}{n}$$
and consequently the wave vector
$$k \to k_0 n ?$$
Is it enough to account for this wavelength change to derive the EM field equations in a homogeneous material of refraction index n?

2. May 11, 2017

### gleem

The equation of a one dimensional traveling wave is
2A/∂x2 = (1/v2)∂2A/∂t2

where v is the velocity of the wave.

Maxwell's wave equation is

2E/∂x2 = (εμ)2E/∂t2

Where ε and μ are the permittivity and permeability of the material respectively

recognizing that this is a traveling wave equation the velocity of the wave would be

v= 1/(εμ)½

in a vacuum v =c = 1/(ε0μ0)½

3. May 11, 2017

### 2sin54

Thank you for the reply. Yes, the velocity changes and consequently both the wavelength and the wave number change as well. So changing the velocity, wavelength and all the functions of these quantities of the EM wave should be sufficient to get the equations of the EM wave in a material, is that correct?

4. May 11, 2017

### gleem

I do not understand your question. The EM wave equation is derived directly for Maxwell's four equations. It relates the spatial variation to the time variation. The solution of the wave equation then provides the velocity. The wavelength and frequency are determined by the source of the wave not the wave equation itself.

5. May 11, 2017

### 2sin54

Sorry for not being clear. I meant to say that I have the equations for the EM wave (in terms of E and B) in a vacuum (n = 1) and I wish to determine the equations for the same wave when it travels through a homogeneous medium (n != 1).

6. May 11, 2017

### gleem

The only difference is that ε0μ0 is replaced by εμ for the particular medium you are interested in remember

μ = Kmμ0 Km = relative permeability

ε= Keε0 = Ke = dielectric constant.(relative permittivity )

7. May 11, 2017

### 2sin54

Thank you. That more or less confirms my thoughts.