Wavlnumber in media

1. Aug 26, 2014

Roodles01

1. The problem statement, all variables and given/known data
Glass has a permeability of ε=2.0 and relative permeability μ=1.
Electromagnetic radiation of angular frequency ω=2.8x109 s-1. travels through the glass. Calculate
a) refractive index

a, b and c are straightforward but d I have questions for

2. Relevant equations
n=√ε
n=c/v
f=ω/2∏
v=ω/k
ω=2∏f
k=n2∏f/v

3. The attempt at a solution

a) refractive index, n=√ε = √2

b) speed of radiation travelling through glass,
n = c/v
so v = c/n
v = 3.0x108 ms-1 / √2
v = 2.12x108 ms-1

c) frequency of radiation, f = ω/2∏
f = 2.8x109 s-1 / 2∏
f = 4.46x106 Hz

now v = ω/k - where k is the wavenumber
so make k the subject
k = ω/v
k = 2.8x109 s-1 / 2.12x108 ms-1
k = 13.2

BUT

ω = 2∏f
and
k=n2∏f/v

so k = (2x2∏x 4.46x106) / 2.12x108 ms-1
k = 0.132

2. Aug 26, 2014

Simon Bridge

The first one.

Note: $y(x,t)=A\sin(kx-\omega t + \delta)$ ... which means that $k=2\pi/\lambda$ ... i.e. inverse wavelength in the medium.

If v is the speed in the medium, then $v=f\lambda \implies c=nf\lambda$ is the speed in a vacuum.

So $k=2n\pi f/c$ ... recognize it?

3. Aug 26, 2014

collinsmark

Don't you mean relative permittivity?

Try that one again. Keep an eye on the exponent. You're a couple orders of magnitude off there.

4. Aug 26, 2014

Roodles01

Yes, I see.
And yes, typo's.

Ta.