Calculate Wavenumber of Radiation Through Glass

  • Thread starter Roodles01
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In summary, the conversation involved calculating various properties of electromagnetic radiation traveling through glass, including the refractive index, speed, frequency, and wavenumber. The correct calculations for each are as follows: a) Refractive index, n = √ε = √2 b) Speed of radiation, v = 3.0x108 ms-1 / √2 = 2.12x108 ms-1 c) Frequency of radiation, f = ω/2π = 2.8x109 s-1 / 2π = 4.46x108 Hz d) Wavenumber of radiation, k = ω/v = 2πf/c = (2x
  • #1
Roodles01
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Homework Statement


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
b) speed of radiation
c) frequency of radiation
d) wavenumber of radiation

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


Homework Equations


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



The Attempt at a Solution



a) refractive index, n=√ε = √2

b) speed of radiation traveling 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

d)wavenumber of radiation
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

So which is right, please?
 
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  • #2
So which is right, please?
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?
 
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  • #3
Roodles01 said:

Homework Statement


Glass has a permeability of ε=2.0
Don't you mean relative permittivity? :biggrin:

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

Try that one again. Keep an eye on the exponent. You're a couple orders of magnitude off there.
 
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  • #4
Yes, I see.
And yes, typo's.

Ta.
 
  • #5



Both of your attempts at calculating the wavenumber are correct. The first calculation (k = ω/v) gives you the wavenumber in units of m-1, while the second calculation (k = n2∏f/v) gives you the wavenumber in units of cm-1. This is because the speed of radiation (v) is given in meters per second, and the frequency (f) is given in hertz. To convert from m-1 to cm-1, you can simply multiply by 100. So in this case, the wavenumber is 13.2 m-1 or 132 cm-1. Both are correct, it just depends on the unit you prefer to use.
 

1. What is a wavenumber and why is it important in radiation through glass?

A wavenumber is a unit of measurement used to describe the wavelength of electromagnetic radiation. It is important in radiation through glass because the wavenumber determines the amount of energy that is transmitted through the glass. A higher wavenumber indicates a shorter wavelength and higher energy radiation, while a lower wavenumber indicates a longer wavelength and lower energy radiation.

2. How do you calculate the wavenumber of radiation through glass?

The wavenumber of radiation through glass can be calculated using the formula wavenumber = 1/wavelength, where the wavelength is measured in units of length (such as nanometers or micrometers). The resulting wavenumber will be in units of inverse length (such as cm^-1 or m^-1).

3. What factors affect the wavenumber of radiation through glass?

The wavenumber of radiation through glass is affected by several factors, including the refractive index of the glass, the thickness of the glass, and the type of radiation (such as visible light, ultraviolet, or infrared). Additionally, the temperature and pressure of the glass can also impact the wavenumber.

4. How does the wavenumber of radiation through glass change with different types of glass?

The wavenumber of radiation through glass can vary depending on the type of glass used. This is because different types of glass have different compositions and refractive indices, which can affect how much energy is transmitted through the glass. For example, borosilicate glass has a lower wavenumber than soda-lime glass due to its higher refractive index.

5. Is there a limit to the wavenumber of radiation that can pass through glass?

Yes, there is a limit to the wavenumber of radiation that can pass through glass. This is known as the cutoff wavenumber and it is determined by the properties of the glass, such as its refractive index and thickness. Above the cutoff wavenumber, the glass will absorb all of the radiation instead of transmitting it through.

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