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

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 11K views
Alt+F4
Messages
305
Reaction score
0
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??
 
Physics news on Phys.org
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
 
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?
 
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?