How Many Polarizing Sheets Are Needed to Rotate Light 90 Degrees?

  • Context:
  • Thread starter Thread starter bberns
  • Start date Start date
  • Tags Tags
    Light Polarization
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 4K views
bberns
Messages
1
Reaction score
0
Question: We want to rotate the direction of polarization of a beam of light through 90 degrees by sending the beam through one or more polarizing sheets. (a) What is the minimum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than 60% of the original intensity?

Relevant Equations: I(n)=I(0)*(cos(90/n))^(2n)

I have had this problem explained to me once before but have had trouble following the logic of the problem and understanding how the direction of the light changes when moving through a polarizing sheet. Any and all help is greatly appreciated!
 
on Phys.org
bberns said:
Question: We want to rotate the direction of polarization of a beam of light through 90 degrees by sending the beam through one or more polarizing sheets. (a) What is the minimum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than 60% of the original intensity?

Relevant Equations: I(n)=I(0)*(cos(90/n))^(2n)

I have had this problem explained to me once before but have had trouble following the logic of the problem and understanding how the direction of the light changes when moving through a polarizing sheet. Any and all help is greatly appreciated!

Hi bberns! Welcome to MHB! (Smile)

When polarized light enters a polarization filter that is at an angle of 90 degrees, no light can get through.
However, if the filter is at an angle of, say, 45 degrees, the amplitude is reduced by a factor of $\cos(45^\circ) = \frac 12 \sqrt 2$.
Since the intensity is related to the square of the amplitude, the intensity is reduced by a factor of $(\cos(45^\circ))^2 = \frac 12$, after which the process can be repeated to ultimately be left with a non-zero intensity at 90 degrees.

The formula shows how much intensity remains after $n$ filters that are at an angle of $90/n$ degrees with respect to each other.