2 Polarizers attenuating a light beam

In summary: Polarized light going through a filter at angle α emerges with intensity I0cosα2. A single photon going through the same filter has the same probability of being detected on the far side. Is this a general rule: intensity reduction factor = single photon detection probability?
  • #1
stefan3423
6
0

Homework Statement


An angle is given between 2 polarizer's (45 degrees), through them light passes (unpolarized than after passing through the first one it polarizes), some of the light its shown on the display. For how much does the angle needs to be increased for the intensity of light to be 2 times less if the starting angle is 45 degrees.
Given Var:
α=45°

Homework Equations


I=I0cos2α

The Attempt at a Solution


The answer in textbook:[/B]
α21=+15°
My work:
ph.jpg

My answer:(Side note I have a little error instead of II2 its I2, everything else is correct).
solution- Page 1.jpg
 

Attachments

  • ph.jpg
    ph.jpg
    23.4 KB · Views: 883
  • solution- Page 1.jpg
    solution- Page 1.jpg
    21 KB · Views: 364
Last edited:
Physics news on Phys.org
  • #2
Hi Stefan3423 and welcome to PF.

If the light intensity before going through the first polarizer is I0, what is it after it passes through it?
 
  • Like
Likes stefan3423
  • #3
stefan3423 said:

Homework Statement


An angle is given between 2 polarizer's (45 degrees), through them light passes (unpolarized than after passing through the first one it polarizes), some of the light its shown on the display. For how much does the angle needs to be increased for the intensity of light to be 2 times less if the starting angle is 45 degrees.
Given Var:
α=45°

Homework Equations


I=I0cos2α

The Attempt at a Solution


The answer in textbook:[/B]
α21=+15°
My work:View attachment 217015
The diagram is ok, but that is not really an attempt at a solution. Please show some thoughts at least.
 
  • #4
haruspex said:
The diagram is ok, but that is not really an attempt at a solution. Please show some thoughts at least.
Well I know that cos of 90 degrees=0 which means no light will go through the crystal and if cos of 0 degrees=1 i.e the crystal cross section will match thus max polarised light will pass through
 
  • #5
kuruman said:
Hi Stefan3423 and welcome to PF.

If the light intensity before going through the first polarizer is I0, what is it after it passes through it?
Another polirised beam of light with max intensity but the secound poliriser is at angle of 45 degrees
 
  • #6
stefan3423 said:
Another polirised beam of light with max intensity but the secound poliriser is at angle of 45 degrees
You did not answer my question. If the intensity of the beam is I0 before the first polarizer, what is it after that polarizer? I am talking about the intensity between P1 and P2 in your drawing. There is no 45 degree angle involved, yet.
 
  • #7
kuruman said:
You did not answer my question. If the intensity of the beam is I0 before the first polarizer, what is it after that polarizer? I am talking about the intensity between P1 and P2 in your drawing. There is no 45 degree angle involved, yet.
It starts with 45 degrees.BTW I solved it, found my errors and I will post my answer now.
 
  • #8
stefan3423 said:
It starts with 45 degrees.BTW I solved it, found my errors and I will post my answer now.
Please do so.
 
  • #9
kuruman said:
You did not answer my question. If the intensity of the beam is I0 before the first polarizer, what is it after that polarizer? I am talking about the intensity between P1 and P2 in your drawing. There is no 45 degree angle involved, yet.
 

Attachments

  • solution- Page 1.jpg
    solution- Page 1.jpg
    21 KB · Views: 502
  • #10
Instead of II2 is I2=I1/2
 
  • #11
Looks good.
 
  • Like
Likes stefan3423
  • #12
Polarized light going through a filter at angle α emerges with intensity I0cosα2. A single photon going through the same filter has the same probability of being detected on the far side. Is this a general rule: intensity reduction factor = single photon detection probability? In which case, the Schroedinger function, where probability = amplitude2, is a kind of "square root wave"?
 

Related to 2 Polarizers attenuating a light beam

What is the purpose of using 2 polarizers to attenuate a light beam?

The purpose of using 2 polarizers is to reduce the intensity of a light beam by selectively allowing certain polarized light waves to pass through while blocking others. This process of attenuation is useful in a variety of scientific experiments and applications.

How do 2 polarizers attenuate a light beam?

When two polarizers are placed in line with each other, the first polarizer will filter out any light waves that are not aligned with its polarization axis. The remaining light will then pass through to the second polarizer, which will further filter out any light waves that are not aligned with its polarization axis. The result is a significantly attenuated light beam.

Can the amount of attenuation be controlled with 2 polarizers?

Yes, the amount of attenuation can be controlled by adjusting the angle between the polarization axes of the two polarizers. The greater the angle, the more light will be blocked, resulting in a higher level of attenuation. This angle can be adjusted by rotating one of the polarizers.

What types of light can be attenuated with 2 polarizers?

2 polarizers can attenuate all types of light, including visible light, infrared light, and ultraviolet light. The effectiveness of attenuation may vary depending on the wavelength and intensity of the light being passed through the polarizers.

Are there any potential limitations or drawbacks to using 2 polarizers to attenuate a light beam?

One potential limitation is that using multiple polarizers can also result in unwanted effects such as color distortion or polarization dependent loss. Additionally, if the polarizers are not aligned properly, the attenuation may not be as effective. Careful consideration and experimentation may be necessary to achieve the desired level of attenuation without any unwanted effects.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
406
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
8
Views
1K
Back
Top