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(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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- Thread starter Alt+F4
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- #1

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(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??

- #2

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(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 cant get the answer either thanks

- #3

Hootenanny

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How far will the wave have travelled in 9 seconds? What is the volume of the cuboid traced by the surface and the distance travelled by the wave?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?

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the wave would have traveled 17.1 Meters in 9 seconds is that right? (1.9*9) = 17.1Hootenanny said:How far will the wave have travelled in 9 seconds? What is the volume of the cuboid traced by the surface and the distance travelled by the wave?

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Hootenanny

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Are you sure about that? How fast does an EM wave travel?Alt+F4 said:the wave would have traveled 17.1 Meters in 9 seconds is that right? (1.9*9) = 17.1

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3*10^8 m/sHootenanny said:Are you sure about that? How fast does an EM wave travel?

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:

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i have no idea on this one, i cant get the ratio at allAlt+F4 said:

(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 cant get the answer either thanks

- #8

Hootenanny

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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

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(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

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do u have to calculate two diffrent numbers and then add them?Alt+F4 said:

(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?

- #11

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help meeeeeee

- #12

Hootenanny

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Note my original Equation above;

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

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

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