How Does the Angle of Polarizing Sheets Affect Transmitted Light Intensity?

  • Thread starter Thread starter phy112
  • Start date Start date
  • Tags Tags
    Intensity Light
AI Thread Summary
The discussion revolves around calculating the intensity of transmitted light through a system of polarizing sheets. When unpolarized light of intensity I0 passes through the first polarizer, its intensity is reduced to 0.5I0. The introduction of a third polarizing sheet at 46° to the first alters the intensity further, requiring the use of the equation I = I0 cos²θ for accurate calculations. The confusion arises regarding the impact of the middle sheet on the overall intensity, particularly when it is removed, which results in zero intensity due to the orthogonal positioning of the first and second sheets. The thread emphasizes breaking down the problem into manageable steps to clarify the effects of each polarizer on light intensity.
phy112
Messages
37
Reaction score
0

Homework Statement



Two sheets of polarizing material are placed with their transmission axes at right angles to each other. A third polarizing sheet is placed between them with its transmission axis at 46° to the axis of the first one.

(a) If unpolarized light of intensity I0 is incident on the system, what is the intensity of the transmitted light?
X Io

(b) What is the intensity of the transmitted light when the middle sheet is removed?
X Io



Homework Equations



I=.5Io
I=Iocos2theta

The Attempt at a Solution



I am confused on what they want! (the answers are in X Io
 
Physics news on Phys.org
so i got b is zero(which is correct). is A the same?
 
No, A is different.

Break it up into single steps. You already know what the first polarizer does to intensity. Each addition polarizer transmission depends on it's angle relative to the previous polarizer.
 
i tried every thing and i can't figure it out. is there any more hints
 
Step 1:
What is the intensity after the 1st polarizer?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Effect of a polarizer/analyzer on partially polarized light
Replies
25
Views
3K
Graphing Power vs Polarizer Angle using Intensity
Replies
32
Views
2K
Why Does Unpolarized Light Lose Half Its Intensity Through a Polaroid?
Replies
4
Views
2K
Intensity of light passing through 3 polarizers
Replies
3
Views
6K
Find angle of polarizing sheet given original & transmitted intensity
Replies
4
Views
5K
Does light intensity change with two polarizing sheets?
Replies
3
Views
2K
Calculating Polarization Intensity: Theta Values of 45 and 25 Degrees Explained
Replies
3
Views
5K
Polarization of Light: Polarizer
Replies
7
Views
2K
Laser Spectroscopy: Calculating Transmitted Intensity
Replies
7
Views
1K
Understanding Malus Law and Polaroid Intensity: Conceptual Doubts
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top