Light intensity in matter

In summary: Emax^2) = .5*(.704)(7.20*10^-3)^2 ≈ 1.41*10^-5In summary, to calculate the intensity of an electromagnetic wave in an insulating magnetic material, modify the formula for intensity in a vacuum by replacing the permittivity and permeability with the values given for the medium. This will result in an intensity of approximately 1.41*10^-5 for an electromagnetic wave with frequency 65.0Hz and an electric field amplitude of 7.20×10−3V/m.
  • #1
sunmaggot
62
4

Homework Statement


An electromagnetic wave with frequency 65.0Hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude 7.20×10−3V/m.

What is the intensity of the wave?

Homework Equations


I = S average = 1/2 ε0 c Emax2

The Attempt at a Solution


the above equation is what I only have from my textbook, and it is for vacuum case. However, in the question, it has dielectric constant and relative permeability. I tried changing c into the light speed in the medium, not correct. I tried using equation from wiki, that is, times refractive index in the above equation, doesn't work. I tried simply ignore the refractive index problem, and doesn't work.
 
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  • #2
Please show your working.
How do you know "it didn't work"?

You book should have shown you a derivation for the formula it gave you... check that you are accounting for all the assumptions correctly.
Your book should include some text about how EM fields are different in a medium.
Have a go deriving the formula for intensity in a magnetic medium.
 
  • #3
Can someone help? I have literally the exact same question, which I already gave up on and have the answer for, but still don't understand.

My book says that the required modifications for the intensity in a medium are "simple." Replace ##\epsilon_0## with the permittivity ##\epsilon##, replace ##\mu_0## with the permeability ##\mu##, and replace ##c## with the speed ##v##.

I replaced ##\mu_0## with the permeability given, and replaced ##c## with the speed found in a previous problem, and still got the problem completely wrong (by many orders of magnitude). I did:
$$(Emax * Bmax)/(2*5.18)$$ and $$(Emax ^2)/(2*5.18*v)$$ [all with the correct substitutions as found in previous problems] and both ended up with the same result, which was wrong.
 
  • #4
If you use I = .5*sqrt( ε0/μ0)(Emax^2) and modify it with the dielectric constant and relative permittivity like so
I = .5*sqrt( ε0*3.64/(μ0*5.18))(Emax^2)
 

What is light intensity in matter?

Light intensity in matter refers to the amount of light that is absorbed, transmitted, or reflected by a material. It is a measure of the brightness of light that is passing through or interacting with a substance.

How is light intensity in matter measured?

Light intensity in matter can be measured using a device called a spectrophotometer, which measures the amount of light that is absorbed or transmitted by a material. It can also be measured using a photometer, which measures the intensity of light emitted from a source.

What factors affect light intensity in matter?

The main factors that affect light intensity in matter include the properties of the material, such as its density, transparency, and color, as well as the wavelength and intensity of the light source. Other factors include the angle of incidence, temperature, and the presence of any impurities or contaminants.

How does light intensity in matter impact our daily lives?

Light intensity in matter plays a crucial role in our daily lives. It allows us to see and perceive colors, shapes, and textures. It also affects the growth and development of plants, the effectiveness of solar panels, and the functioning of optical devices such as cameras and microscopes.

What are some practical applications of understanding light intensity in matter?

Understanding light intensity in matter has many practical applications in various fields. For example, in medicine, it is used to diagnose diseases and monitor the effectiveness of treatments. In materials science, it is used to study the properties and behavior of different materials. In agriculture, it is used to optimize plant growth and yield. In environmental science, it is used to monitor air and water quality. Additionally, it has numerous applications in industries such as photography, printing, and electronics.

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