# Solution to Exercise 5.1-3 Polarization

| 1. question:: A light beam with intensity I0 passes through one ideal polarizer.
|  | 1. How does the intensity relate to the electrical field strength?
|  | Intensity neasures the energy or better power flux contained in the light. it is propotional to the square of the electrical field strength
|  | 2. The incoming ("input") light beam is unpolarized. How large is the intensity at the output?
|  | It's obviously I0/2. Considering this picture we can decompose the light beam in two fully polarized beams, each having the intensity I0/2. The polarizer takes out one of the beams and I0/2 remains.
|  | 3. Does this intensity change if you rotate the polarizer around the axis coinciding with the propagation direction of the light = optical axis?
|  | No.
|  | 4. The incoming light beam is 100 % linearly polarized. How large is the intensity on the output as a function of the angle between polarization direction of the light and polarizing direction of the polarizer.
|  | The intensity must vary betwen 100 % and 0 % of the incoming intensity for an angle of a = 0o and 90o , respectively. For an arbitrary angle a we have a field strength E(a) = E0 · cosa; the transmitted intensity then scales with (cosa)2 between the extremes.
| 2. question:A light beam with intensity I0 first passes through one ideal polarizer, and then through a second one. Both polarizers can be rotated freely around the optical axis.
|  | 1. The light beam is unpolarized. How large is the intensity on the output if both ideal polarizers are in parallel?
|  | It's still I0/2 because two polarizers in parallel behave just like one.
|  | 2. The light beam is unpolarized. How large is the intensity on the output if the ideal polarizers are "crossed", i.e. their polarization directions are at right angles?
|  | The intensity is zero.
|  | 3. The light beam is 100 % linearly polarized. How large is the intensity of the output as a function of the variable angle a between the two polarizing directions of the polarizers and the fixed angle b between the polarization direction of the light and the first polarizer it encounters? Note that in this case you rotate the second polarizer.
|  | After passing through the first polarizer, the intensity is I0 · (cosb)2. After passing through the second polarizer we have I = I0 · (cosb)2 · (cosa)2
|  | 4. Does the result for the question above change if you rotate the first polarizer and keep the second one at the fixed angle b?
|  | No.
|  | 5. Is for all of the above the direction of the light paths always reversible as stated before?
|  | Tricky. Just reversing arrowheads obviously doesn't work. You can't get a higher intensity from a lower one as you would if you just reverse the sign of the Poyntig vector as shown below. Using a mirror symmetry works but this trival.
Starting situation. | Reversing vector signs does not work.
Mirroring works.
| 3. question: Now consider a system with two fixed crossed polarizers and a third one that can be rotated in between the two crossed ones.
|  | 1. The incoming light beam is unpolarized. How large is the intensity of the output as a function of the variable angle a between the first (fixed) and the third polarizer that can be rotated?
|  | After the first polarizer, the intensity is I0/2; it decreases with (cosa)2 behind the third polarizer that can be rotated . The second (fixed and crossed with respect to the first one) polarizer transmits components with a (sina)2 dependence (make a simple drawing and do the geometry!) so all together we have for the output
I (a) = | I0 2 | · | (cosa)2 · (sina)2
--- | --- | --- | ---
| I (a = 0o) = 0
I (a = 45o) = maximal value = (I0/2) · 0,25
I (a = 90o) = 0
|  | 2. The incoming light beam is 100 % linearly polarized. How large is the intensity on the output as a function of the variable angle a between the first (fixed) and the third polarizer (can be rotated) considering that the angle b between the incoming light polarization and the polarization direction of the first polarizer is fixed at a value b?
|  | As above except that the intensity after the first polarizer is now I0 · (cosb)2

With frame

Exercise 5.1.3 Polarization

© H. Föll (Advanced Materials B, part 1 - script)