How Is Snell's Law Derived Using Boundary Conditions and Wave Equations?

[helpcometk]

Homework Statement

The electric field of an electromagnetic wave traveling in the dielectric can be written as
E = (E0) e(iωt−ikx)
where E0 is a constant vector and |k| = [ω(ε)^1/2]/c .

Consider an electromagnetic wave traveling in vacuum towards the dielectric at an angle θ
to the x-axis with its electric field polarised in the x, y plane. When this wave is incident
on the dielectric a scattered wave and a transmitted wave are produced.
By imposing the relevant boundary conditions on the electric field across the boundary, or
otherwise, derive Snell’s law.

Homework Equations

Snell's law : (sinθ1)(n1)=(sinθ2)(n2)
Where:
θ1 is the angle of incidence and θ2 is the angle of reflection,measured with respect to the normal.
n is the refractive index of the material
n1,2 = velocity of light in a vacuum / velocity of light in medium

ω=2πu=ku
where :
k is the wave number defined as:
λ=2π/k

ω is the angular frequency
u is the frequency

The Attempt at a Solution

 

[henry_m]

Hi,

What does the electric field of the incident wave look like? What about the reflected and transmitted waves? What are the boundary conditions?

 

[helpcometk]

So you actually say that the exponents of reflected and incident waves are equal with the exponent of the transmitted wave.which means xk1=xk2. ?

 

[henry_m]

I'm not quite sure what you mean. My point was that you should start by explicitly writing down what the electric field of an incoming wave as described in the problem would be, and write down what the boundary conditions are at the interface. You should then find that the forms of the reflected and transmitted waves are heavily constrained by the boundary conditions.

 

[paranormal]

To derive Snell's law, we must first consider the boundary conditions at the interface between the vacuum and the dielectric material. At this interface, the incident wave will be partially reflected and partially transmitted, resulting in both a reflected and transmitted wave. The electric field at the interface can be described as:

Ei = E0e^(iωt-ikx)

where E0 is the amplitude of the incident wave, ω is the angular frequency, and k is the wave number.

The reflected wave can be described as:

Er = E0re^(iωt+ikx)

where E0r is the amplitude of the reflected wave and k is the wave number.

The transmitted wave can be described as:

Et = E0te^(iωt-ikx)

where E0t is the amplitude of the transmitted wave and k is the wave number.

By applying the boundary conditions at the interface, we can equate the electric field at the interface for the incident wave, reflected wave, and transmitted wave:

Ei + Er = Et

This can be rewritten as:

E0e^(iωt-ikx) + E0re^(iωt+ikx) = E0te^(iωt-ikx)

Simplifying and rearranging, we get:

(E0 + E0r)e^(iωt-ikx) = E0te^(iωt-ikx)

Dividing both sides by E0e^(iωt-ikx), we get:

1 + r = t

where r = E0r/E0 and t = E0t/E0 are the reflection and transmission coefficients, respectively.

Next, we can use the definition of the refractive index to relate the wave number k in vacuum to the wave number k' in the dielectric material:

k' = k/n

where n is the refractive index of the dielectric material.

Using the definition of the wave number, we can rewrite this as:

2π/λ = (2π/λ')/n

where λ and λ' are the wavelengths in vacuum and the dielectric material, respectively.

Simplifying, we get:

n = λ/λ'

Substituting this into our previous equation, we get:

1 + r = t

We can also use the definition of the angle of incidence and Snell's law to relate the incident, reflected, and transmitted angles: