Index of refraction from electromagnetic wave equation in liquid

[bmb2009]

ElectroMagnetic Wave In Liquid.. Help with Index of Refraction??

Homework Statement

Consider the propagation of an electromagnetic wave in some liquid. The y-component of its electric field is given by Ey(x,t)=4.3 x104 V/m sin(5.40E-2m-1x + 6.00E+0 x106 s-1 t). How large is the index of refraction of that liquid?

Homework Equations

The Attempt at a Solution

Honestly had no idea how to even start this problem? Any help would be great. Thanks!

 

[Ibix]

Your equation is quite confusing. Did you mean:
E_y(x,t)=4.3x10⁴ V/m sin(5.40x10^-2m^-1x + 6.00x10⁶ s^-1 t)

What you have is an equation describing the electric field of the EM wave at point x and time t.

If you freeze time at t=0, the equation simplifies somewhat. Can you read off the wavelength?

If you stand at the point x=0, the equation also simplifies somewhat. Can you read off the frequency?

That should give you enough information to derive the refractive index. Post your working if you get stuck.

 

[bmb2009]

Yes sorry the equation you interpreted was what I meant... I felt like I know what I am doing but still can't get the answer. Here's what i tried. since the speed of light velocity in a material I'll call v, equals the wavelength in the material times the frequency i.e. v=wavelength*frequency and since the index of refraction, n, equals c/v...If I can find the wavelength and the frequency I should be able to find the v and in turn the index of refraction. I know from my textbook that E(x,y)=E_0(sin(2*Pi(x/wavlength - frequency*t))) so I should just have to factor out a 2Pi from the terms inside the sine correct? I tried this and it did not work... Help :/

 

[Ibix]

Your methodology seems right. Perhaps you made an arithmetic error? Post your wavelength, frequency, velocity and index, and I can take a look.

By the way - I presume the wave is moving in the x-direction? If not, your wavelength will be wrong.

 

[bmb2009]

I assume it's in the x direction because the question is posted verbatim from the textbook.
...
My wavelength: factor 2*pi from the term (x/.054m) to get x/(.054*2pi) so my wavlength =
.33929 m.
MY frequency: factor 2*pi from the term (6x10^6) = (6x10^6)/(2*pi) so my frequency = 3.23998x10^5

v=1.099x10^5 ... this is where I am confused because to get a reasonable number for the index of refraction the velocity will have to be some number x10^7 correct?

If you could run the numbers and post what you get I'd be very grateful. Thanks

 

[bmb2009]

Nevermind I just re did the problem and got the right answer... thank you for your help though i really appreciate it!