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

AI Thread Summary
An electromagnetic wave with a frequency of 5.7 x 10^14 Hz has an average total energy density of 4.6 x 10^-6 J/m3. A flat surface of 1.9 m2, perpendicular to the wave's propagation, receives approximately 23,598 J of energy over 9 seconds. The wave travels at the speed of light, approximately 3 x 10^8 m/s, covering a distance of about 5.13 x 10^9 meters in that time. The discussion also touches on calculating the electric field amplitude after passing through polarizers, with ratios provided for intensity changes. The calculations involve using the intensity ratios and electric field relationships, emphasizing the importance of understanding the underlying physics principles.
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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.

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For your next question it may be useful to note that;

\frac{I}{I_{0}} = \cos^2\theta

and

E = E_{0}\cos\theta

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 alot, 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 alot, 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;

\frac{I}{I_{0}} = \cos^2\theta
 
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