An Electromagnetic wave goes from air into a medium....

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Homework Help Overview

The discussion revolves around the behavior of an electromagnetic (EM) wave transitioning from air into a different medium. The problem involves analyzing the electric field equations of the waves in both media and determining the relationship between their relative permittivities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the transformation of the electric field equations and the implications of wave number and angular velocity changes. Questions are raised about the factors that change when an EM wave enters a new medium, particularly regarding the refractive index and the role of the medium's magnetic properties.

Discussion Status

Some participants have provided guidance on the relationships between wave properties and the implications of the medium's characteristics. There is an acknowledgment of potential errors in reasoning, and the discussion is exploring multiple interpretations of the problem without reaching a consensus.

Contextual Notes

Participants note the relevance of the medium being non-magnetic and question the assumptions regarding the relative permittivities of air and the new medium. There is also a mention of the need for clarity on the definitions and relationships involved in the equations presented.

palkia
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Homework Statement


An EM wave from air enters a medium.
The electric fields are

--> ^
E1 = E01 cos(2πv(z/c-t) x
--> ^
E2 =E02cos(k(2z-ct)x in a medium ,where the wave number k and frequency v refer to their value in air.the medium is non-magnetic.If εr1 and εr2 refer to relative permittivities of air and medium respectively,which of the following is correct

(a)εr1/εr2=1/4
(b))εr1/εr2=1/2
(c))εr1/εr2=4
(d))εr1/εr2=2

Homework Equations



c=ω/k =v [/B]λ=1/ √μ0ε0

The Attempt at a Solution



So I first tried to bring the equation in the form-E0sin(ωt-kz)

Using the relations I found
E1=E01cos(kz-wt)
E2=E02cos(2kz-wt)

So the wave number becomes double and angular velocity remains same

So ω/k= 1/ √μ0ε0
Using this I get εr1/εr2=4

Did I do this correctly?

Can somebody also tell me what factors change when electromagnetic wave changes medium and what's up with refractive index change with it...is it because the velocit changes.I am kinda new to EM waves so that's why I am asking such dufus questions XD

What does non-magnetic field medium do int his question
 
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palkia said:
ε

E1 = E01 cos(2πv(z/c-t) x
--> ^
E2 =E02cos(k(2z-ct)x
I assume x is a unit vector indicating the direction of the E field vector?
Using this I get εr1/εr2=4
Did I do this correctly?
you didn't show all your in-between steps but the answer is correct.
Can somebody also tell me what factors change when electromagnetic wave changes medium and what's up with refractive index change with it...is it because the velocity changes.I am kinda new to EM waves so that's why I am asking such dufus questions XD
What does non-magnetic field medium do int his question
If you look at your expression ω/k= 1/√μ0ε0 you can show that the material's magnetic property (i.e. its permeability μ) is irrelevant providing it is the same for both waves E1 and E2. So it doesn't have to be non-magnetic per se.

Remember that the propagation speed ∝ 1/√ε or 1/n and k = 2π/λ and λf = v, f = frequency, v = speed.
 
rude man said:
I assume x is a unit vector indicating the direction of the E field vector?you didn't show all your in-between steps but the answer is correct.If you look at your expression ω/k= 1/√μ0ε0 you can show that the material's magnetic property (i.e. its permeability μ) is irrelevant providing it is the same for both waves E1 and E2. So it doesn't have to be non-magnetic per se.

Remember that the propagation speed ∝ 1/√ε or 1/n and k = 2π/λ and λf = v, f = frequency, v = speed.
@rude man ## \epsilon_{r1} ## is the air. Please look at his answer again. I believe it is incorrect.
 
Charles Link said:
@rude man ## \epsilon_{r1} ## is the air. Please look at his answer again. I believe it is incorrect.
I agree, I goofed.
E2 has the lower lambda and so of course the higher permittivity. Sorry, OP.
 
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