An unpolarized beam of intensity I o = 546 W/m 2 travels in the positive z-direc

Question

An unpolarized beam of intensity Io = 546 W/m2 travels in the positive z-direction and is incident from the left on a series of three linear polarizers as shown. The transmission axis of the first polarizer is aligned with the y-axis. The following polarizers make angles of 2 = 69 degrees, and 3 = 29 degrees with the positive x axis.

1)

What is I2, the intensity of the beam immediately following the second polarizer?

W/m2

Your submissions:

111.65

Computed value:

111.65

Submitted:

Tuesday, May 10 at 6:28 PM

Feedback:

2)

What is Ifinal, the intensity of the beam immediately following the third polarizer?

W/m2

Your submissions:

111.65

Computed value:

111.65

Submitted:

Tuesday, May 10 at 6:28 PM

Feedback:

3)

The positions of the second and third polarizers are now interchanged. What is Ifinal,new, the intensity of the beam after passing through the new arrangement of the three polarizers.

W/m2

4)

Suppose we now move our incident unpolarized beam so that it is incident from the right. That is, the beam is now incident upon the polarizer whose transmission axis makes an angle 2 with the x-axis and exits the last polarizer (with transmission axis aligned with the y-axis) with intensity Ifinal,RL. How does Ifinal,RL compare with Ifinal,new?

Ifinal,RL < Ifinal,new

Ifinal,RL = Ifinal,new

Ifinal,RL > Ifinal,new

5)

Suppose we now rotate the last polarizer by 90 degrees so that its transmission axis becomes aligned with the x-axis. With the beam still incident from the right (on the polarizer whose transmission axis makes an angle 2 with the x-axis), what is the Ifinal,RLnew, the intensity of the beam immediately following the last polarizer (transmission axis aligned with the x-axis)?

W/m2

Explanation / Answer

(1) The intensity of the beam immediately following the second polarizer which will be given as :

using a formula, we have

I2 = (I0 / 2) cos2 (90 - 690)

I2 = [(546 W/m2) / 2] cos2 210

I2 = 237.9 W/m2

(2) The intensity of the beam immediately following the third polarizer which will be given as :

using a formula, we have

I3 = I2 cos2 (690 - 290)

I3 = (237.9 W/m2) cos2 400

I3 = 139.5 W/m2

(3) The positions of the second and third polarizers are now interchanged.

The intensity of the beam after passing through the new arrangement of three polarizers which will be given as :

using a formula, we have

Ifinal,new = I2 cos2 (690 - 290) = (I0 / 2) cos2 (90 - 290) cos2 (690 - 290)

Ifinal,new = [(546 W/m2) / 2] cos2 610 cos2 400

Ifinal,new = 37.6 W/m2

(4) Suppose we now move our incident unpolarized beam so that it is incident from the right.

Ifinal,RL = Ifinal,new

Since, the incoming light is unpolarized we run into the same sequence of intensity loss as we would from the other direction. The difference angles between filters 3 and 2 & between 2 and 1 is the same as those coming from the other direction and the loss of half intensity for first filter due to the unpolarized light is the same.

(5) The Ifinal,RLnew, the intensity of the beam immediately following the last polarizer (transmission axis aligned with the x-axis) which is given as :

using the malu's law, we have

Ifinal,RLnew = (I0 / 2) cos2 290 cos2 690

Ifinal,RLnew = [(546 W/m2) / 2] (0.7649) (0.1284)

Ifinal,RLnew = 26.8 W/m2

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