• Support PF! Buy your school textbooks, materials and every day products Here!

More optics

1. *A beam of light that is a mixture of plane polarized and unpolarized light is incident on a sheet of polaroid. The trasmitted intensity varies as the sheet is rotated, having a maximum when the transmission axis is horizontal, and a minimum when it's vertical. If the maximum intensity is three times the minimum, what fraction of the intensity of the original beam is plane polarized, and what is it's direction of polarization?




2. Homework Equations
I=Imax*cos^2*theta
(polarized light passing through analyzer)


3. The Attempt at a Solution

Basically I have no idea with this one, except I'm pretty sure the answer to the last part is that its' direction of polarization is horizontal.
I also just guessed and played around with the fact that with an ideal polarizer the intensity of the transmitted light is half the incident unpolarized light, so I played around with:
Imin= I/2
Imax= 3*Imin
= 3*I/2
= 3/2*I
But that can't be right because its not a proper fraction..
Any help very much appreciated..
 

Answers and Replies

alphysicist
Homework Helper
2,238
1
Hi planesinspace,

I think you need to keep track of what happens to the unpolarized and polarized light separately. Before the light passes through the polarizer, let [itex]I_1[/itex] be the intensity of the unpolarized light, and [itex]I_2[/itex] be the intensity of the unpolarized light. So before the light goes through the polarizer the total intensity is [itex]I_1+I_2[/itex].

When the transmitted light is at a maximum, what is the transmitted intensity? What is it when it at a minimum? How are these two related? Once you have those conditions you can figure out how the original intensities are realated.
 
Hi, thanks for your help.

Oh ok, well I have so far
I(total)= I1 +I2
Where I1 is polarised light, and I2 is unpolarized.
Then, when the light is at a maximum, hence, there is light that is both polarized and unpolarized, the total intensity = 3*Imin = I1 + I2
and when its minimum, only unpolarized light is getting through, hence there is no I1, and Imin = I2 / 2 (because half the transmitted intensity continues, although i read that is only for ideal polaroids, so once again I'm not sure if I is I/2 or I..)
 
alphysicist
Homework Helper
2,238
1
For the unpolarized light, after it passes through a polarizer, its intensity is half what it was before. So your minimum expression [itex]I_{\rm min}[/itex] is correct.

However, you haven't taken that into accout for the maximum. At the maximum intensity, all of the polarized light's intensity passes through, but still only half of the unpolarized light's intensity.
 
So I have Imax= I1 + I2/2 = 3*Imin = 3/2*I2
So when determining what fraction of the total that is, thats where I get confused.
I1 + I2 = I total,
So can i say Itotal = I max, sorry for this confusion.
 
alphysicist
Homework Helper
2,238
1
They want the fraction of the initial beam intensity that was plane polarized. This means they want:

[tex]\frac{I_1}{I_{\rm total}}[/tex]

You already have

[tex]I_1+I_2=I_{\rm total}[/tex]

and you also found:

[tex]I_1 +\frac{1}{2}I_2 = 3\left(\frac{1}{2}I_2\right)[/tex]

These last two can give you a numeric value for the fraction you are looking for after some algebraic cancellation. What do you get?
 
Ok well i got I1 = I2 , which means Itotal = t*I2 ,
I1/Itotal = I2/ 2*I2, so a half?
 

Related Threads for: More optics

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
10
Views
2K
Replies
3
Views
15K
Replies
3
Views
802
Replies
1
Views
140
  • Last Post
Replies
3
Views
520
  • Last Post
Replies
2
Views
573
Top