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

Click For Summary
SUMMARY

An electromagnetic (EM) wave transitioning from air to a medium demonstrates a change in relative permittivity, specifically εr1/εr2=4, as derived from the equations E1 = E01 cos(2πv(z/c-t)) and E2 = E02 cos(k(2z-ct)). The wave number doubles while the angular velocity remains constant, indicating that the medium's properties affect wave propagation. The discussion clarifies that the magnetic properties of the medium do not influence the relationship between the waves, as long as they remain consistent across both mediums.

PREREQUISITES
  • Understanding of electromagnetic wave equations
  • Familiarity with relative permittivity (εr) concepts
  • Knowledge of wave number (k) and angular velocity (ω)
  • Basic principles of wave propagation in different media
NEXT STEPS
  • Study the relationship between refractive index and relative permittivity
  • Learn about the implications of wave propagation in non-magnetic media
  • Explore the derivation of wave equations in different mediums
  • Investigate the effects of frequency on electromagnetic wave behavior
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, wave propagation, and materials science. This discussion is beneficial for anyone seeking to understand the behavior of electromagnetic waves as they transition between different media.

palkia
Messages
52
Reaction score
2

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
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Charles Link

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
12K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
11K
  • · Replies 2 ·
Replies
2
Views
7K