Polarization disk

  • Thread starter Will
  • Start date
  • #1
Will
I am studying for our final, and there is this problem that is confusing me.

There are three disks, with transmission axis of the 1st being 10 deg. from the vertical, the 2nd at an angle theta from the vertical, and the 3rd at an angle 120 deg from the vertical. Unpolarized light, I(0) is incident on the first disk,. We need to find the angle theta for 2nd disk so that transmitied light from 3rd disk is of max intensity.
So I know that the unpolarized light automatically loses 1/2 of its intensity; I(1)=I(0)/2 and that the intensity of I(3) is given by [I(0)/2]cos^2(theta-10)cos^2(120-theta). I think the formula is correct. So how should I go about finding the max? Should I take the derivative of this function and find the zeros, or is there a better way?
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
Hi,

You do not need to take a derivative. The outgoing beam of the first polarizer is the incoming beam for the second, and so on. The I0 for each polarizer is the incoming beam for that polarizer. So, your I0 for the second is simply your answer for the first.
 
  • #3
Will
Will

Still don't quite get it. I(1) emegerges out with a polarized at an angle 10 degrees to the vertical. It then goes through disk 2, which makes a variable angle( the one we need to solve for), and emerges as I(2) with intensity I(1)cos^2(theta-10), right? Then it goes thru disk 3 and emerges with I(3)=I(2)cos^2(120-theta.
If I set theta equal to 10 degrees, I get maximum transmission of I(2). But when it gose thru I(3), cos^2(120-10) is small, so I get a low transmission. So I think I need to find an angle in the somewhere in between 10 and 120 dgrees. Perhaps I was unclear on my wording of the problem, the axis of the first and third disk are fixed , at 10 and 120 respectively, and the angle of the second is variable.

ps how do you make subscripts, and for that matter "pretty print" equations?
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8


Originally posted by Will
I(1) emegerges out with a polarized at an angle 10 degrees to the vertical. It then goes through disk 2, which makes a variable angle( the one we need to solve for), and emerges as I(2) with intensity I(1)cos^2(theta-10), right?
Right.

Then it goes thru disk 3 and emerges with I(3)=I(2)cos^2(120-theta.
If I set theta equal to 10 degrees, I get maximum transmission of I(2). But when it gose thru I(3), cos^2(120-10) is small, so I get a low transmission. So I think I need to find an angle in the somewhere in between 10 and 120 dgrees. Perhaps I was unclear on my wording of the problem, the axis of the first and third disk are fixed , at 10 and 120 respectively, and the angle of the second is variable.
Oops...I made a mistake. I misread your problem, and you were right initially: You do need to take a derivative. Either that, or graph the function to find the peak.

You have:

I2(θ)=I0cos2(θ-10o)

and

I3(θ)=I2(θ)cos2(120o-θ)

which you should rewrite as:

I3(θ)=I0cos2(θ-10o)cos2(120o-θ)

ps how do you make subscripts,
Example: Type

I[ sub ]0[ /sub ]

without the spaces to get:

I0

Replace sub with sup for superscripts.

and for that matter "pretty print" equations?
Look at the Announcement at the top of all the science forums:

Howto: Making Math Symbols.

You can also get math symbols by clicking "Get More" on the smiley menu, but they do not look as good as the ones in the Announcement.
 

Related Threads on Polarization disk

Replies
0
Views
1K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
2
Views
11K
  • Last Post
Replies
5
Views
5K
Replies
5
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
2K
Top