How much energy does a flat surface receive from an electromagnetic wave?

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Homework Help Overview

The discussion revolves around the energy received by a flat surface from an electromagnetic wave, specifically focusing on calculations involving energy density, surface area, and time. The context includes concepts from electromagnetism and wave propagation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to calculate the energy received by a surface based on given energy density and dimensions. Questions arise regarding the distance traveled by the wave and the implications of the calculations. Some participants also discuss the relationship between intensity and electric field amplitude after passing through polarizers.

Discussion Status

The discussion is active with participants questioning their calculations and assumptions. Some guidance has been offered regarding the relationships between intensity, electric field amplitude, and the effects of polarizers. There is a mix of interpretations and calculations being explored without a clear consensus.

Contextual Notes

Participants are working under constraints related to homework guidelines, which may limit the information they can use or the methods they can apply. There is also a focus on ensuring the correct application of formulas related to electromagnetic waves and intensity ratios.

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An electromagnatic wave is traveling in vacuum with frequency 5.7 x 1014 Hz. The wave has average total energy density of 4.6 x 10-6 J/m3.


(e) How much energy does a 1.9 m2 flat surface (perpendicular to the wave propagation direction) receive in 9 s?

E = 23598 J, how do i get that number?


4.6*10^-6 = ATC??
 
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I1 / I0 = .871572

(b) Calculate the electric field amplitude E1 of the light after it has passed through the first polarizer. Express your answer as a fraction of the electric field amplitude E0 of the initial beam.

E1 / E0 = .933


I = c*e0*E2

I can't get the answer either thanks
 
Alt+F4 said:
An electromagnatic wave is traveling in vacuum with frequency 5.7 x 1014 Hz. The wave has average total energy density of 4.6 x 10-6 J/m3.


(e) How much energy does a 1.9 m2 flat surface (perpendicular to the wave propagation direction) receive in 9 s?

E = 23598 J, how do i get that number?
How far will the wave have traveled in 9 seconds? What is the volume of the cuboid traced by the surface and the distance traveled by the wave?
 
Hootenanny said:
How far will the wave have traveled in 9 seconds? What is the volume of the cuboid traced by the surface and the distance traveled by the wave?
the wave would have traveled 17.1 Meters in 9 seconds is that right? (1.9*9) = 17.1
 
Alt+F4 said:
the wave would have traveled 17.1 Meters in 9 seconds is that right? (1.9*9) = 17.1
Are you sure about that? How fast does an EM wave travel?
 
Hootenanny said:
Are you sure about that? How fast does an EM wave travel?
3*10^8 m/s

So 3*10^8 * 1.9 = 5.7E8 meters in a second so 5.13E9 meters in 9 seconds

Ok Got it thanks
 
Last edited:
Alt+F4 said:
I1 / I0 = .871572

(b) Calculate the electric field amplitude E1 of the light after it has passed through the first polarizer. Express your answer as a fraction of the electric field amplitude E0 of the initial beam.

E1 / E0 = .933


I = c*e0*E2

I can't get the answer either thanks
i have no idea on this one, i can't get the ratio at all
 
Alt+F4 said:
3*10^8 m/s

So 3*10^8 * 1.9 = 5.7E8 meters in a second so 5.13E9 meters in 9 seconds

Ok Got it thanks

Careful, your dealing with a volume there not a length, never the less your answer should be correct.

[tex]\hline[/tex]
For your next question it may be useful to note that;

[tex]\frac{I}{I_{0}} = \cos^2\theta[/tex]

and

[tex]E = E_{0}\cos\theta[/tex]

Although it is useful to remember that The ratio of the intensities is equal to the square of the rms of the Electric field.
 
Thanks a lot, one last question


(c) The second polarizer is set at various angles within the range q 2 = 0 to 90°. Calculate the intensity of the light after it has passed through the second polarizer for the following values of q 2. Express each answer as a fraction of I1.



At q 2 = 22°: I2 / I1 =




So i am guessin i need to find final intensity which would be


S = S0 ((cos theta)^2)^2

is that right?
 
  • #10
Alt+F4 said:
Thanks a lot, one last question


(c) The second polarizer is set at various angles within the range q 2 = 0 to 90°. Calculate the intensity of the light after it has passed through the second polarizer for the following values of q 2. Express each answer as a fraction of I1.



At q 2 = 22°: I2 / I1 =




So i am guessin i need to find final intensity which would be


S = S0 ((cos theta)^2)^2

is that right?
do u have to calculate two diffrent numbers and then add them?
 
  • #11
help meeeeeee
 
  • #12
Note my original Equation above;

[tex]\frac{I}{I_{0}} = \cos^2\theta[/tex]
 

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