Changing X-Polarized Light to Y-Polarization with Lossless Polarizers

[UrbanXrisis]

change x-polarized light to y-polarized light by suing lossless polarizers

Using I_{pol}=Icos^2 \theta I would first use a polarizer with theta=45 to change x polarized light 45 degrees without losing intenisity. Then I wil using a polarizer with theta=135 to change the polarized light y-polarization without losing intensity. is this the correct way of approching this problem?

 

[Galileo]

The idea is good (I think, but I don't know how you define your angles theta). If the polarization direction of the polarizer is at some nonzero angle wrt the polarization direction of the light you always have some loss of intensity, since the E-field component of the light-wave perpendicular to the polarization direction of the polarizer is blocked out.

 

[UrbanXrisis]

say the light is polarized vertically like this:


| | | | | |

then the polarizers that are like this:

\ \ \ \ \

putting the two together, the angle is between the two like this:

\|


would what I did still be corrent according to this definition of theta?

 

[berkeman]

I think what you did is correct, but I would have called the two polarizer angles 45 degrees and 90 degrees. The x-polarization direction is at zero degrees, and the final y-polarization direction is at 90 degrees.

 

[Galileo]

Yes, the setup will give you linearly polarized light in the y-direction. But the intensity is not I_0\cos(\theta).
The light intensity drops twice. Once after the first polarizer and again after the second.

 

[UrbanXrisis]

would a 45 degree (wrt x-axis) polarizer and then a 90 degree (wrt x-axis) polarizer work? since, it it would be...

I'=I cos(45)^2

which would lead to I'/2 (this is polarized 45 degree to x axis)

and then applying the 90 degree (wrt the x axis) polarizer, which makes a 45 degree angle to the light:

I''=I/2 cos(45)^2
I''=I/4, which is 90 degrees wrt to the x, which makes it y polarized

does this work?

 

[Doc Al]

Yes, that would work as you describe.