Two linear polarizers in series

In summary, the optical power of a HeNe laser is 5.0mW and the wavelength is 633nm. The emitted light is linearly polarized. When the first polarizer is removed, the detected power drops to zero. At what angle with respect to the direction of polarization of the laser is the polarization axis of the first polarizer? Using equation (2), the intensity (and power) is proportional to the square of the amplitude of the electric field component of the wave. The desired angle is therefore 31.72°.
  • #1
TheSodesa
224
7

Homework Statement


The optical power of a HeNe -laser is ##P_0 = 5.0mW## and the wavelength ##\lambda = 633nm##. The emitted light is linearly polarized. As the laser beam travels through two in-series -polarizers, the power detected behind the second polarizer ##P_2 = 1mW## . If the first polarizer is removed, the detected power drops to zero.

At what angle with respect to the direction of polarization of the laser is the polarization axis of the first polarizer?

Homework Equations


The power of a harmonic wave:
\begin{equation}
P = \frac{1}{2}\mu \omega ^2 A^2 v = \frac{1}{2} \mu (2\pi f)^2 A^2 v = \frac{1}{2} \mu (2\pi \frac{v}{\lambda})^2 A^2 v,
\end{equation}
where ##\mu## is the linear density of the medium of propagation, ##\omega## is the angular frequency, ##A## is the amplitude of the wave and ##v## its velocity.

Sine of a double angle:
\begin{equation}
sin(2\theta) = 2sin(\theta)cos(\theta)
\end{equation}

The Attempt at a Solution



This question was on today's exam, and I'm pretty sure I messed up. I would like some feedback if at all possible.

What I did was assume that only the projection of the electric field component of the wave onto the polarization axis of the first (and second) polarizer passes through, meaning the electric field component of the wave after the first polarizer
\begin{align*}
E_1
&= E_0 cos(\theta),
\end{align*}
where ##\theta## is the angle we want to find out. As the light passes through the second polarizer, the effect is similar:
\begin{align*}
E_2
&= E_1 cos(\phi)\\
&= E_0 cos(\theta)cos(\phi),
\end{align*}
where ##\phi## is the angle between the axes of polarization of the first and the second polarizers. Since we know that no light passes through the second polarizer if the first one is removed, we know that the projection of the electric field component of the wave onto the axis of polarization of the second polarizer must be zero, and thus they must be at a right angle with respect to each other, meaning that
\begin{align*}
\phi + \theta = 90^{\circ} &\iff \phi = 90^{\circ}-\theta.
\end{align*}
Therefore
\begin{align*}
E_2
&= E_0 cos(\theta)cos(90^{\circ}-\theta)\\
&= E_0 cos(\theta)sin(\theta)|| \text{ Using equation (2)}\\
&= \frac{1}{2}E_0 sin(2\theta)
\end{align*}
The next thing I did was to look at equation (1) and say that the intensity (and power) is proportional (equal) to the square of the amplitude of the electric field component of the wave. I did this assumption, because I didn't have the linear density available to me (what would that be in this context, even?). Regardless, if this is true, then
\begin{align*}
P_2
&= E_2^2\\
&= (\frac{1}{2}E_0 sin(2\theta))^2\\
&= \frac{1}{4}E_0^2 sin^2(2\theta)\\
&= \frac{1}{4}P_0 sin^2(2\theta)
\end{align*}
Solving this for ##sin^2(2\theta)## yields
\begin{align*}
sin^2(2\theta)
&= \frac{4P_2}{P_0}
= \frac{4}{5} \text{ (Since } P_0 = 5mW \text{ and } P_2 = 1mW \text{)}\\
\iff\\
sin(2\theta)
&= \pm \sqrt{\frac{4}{5}}
= \pm \frac{2}{\sqrt{5}}\\
\iff\\
2\theta &= arcsin(\pm \frac{2}{\sqrt{5}}) = \pm 63.43^{\circ}
\end{align*}
and the desired angle is therefore
\begin{align*}
\theta &\approx \pm 31.72^{\circ}
\end{align*}
My hunch is this is wrong, mainly because some people I talked to got a different value for ##\theta## (something like ##53^{\circ}##). My guess is that I made a mistake when I made the assumption about the power being equal to the amplitude of the electric field, but I don't really know what else I could have done there...
 
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  • #2
Have you considered that there might be two valid angles and that they might be complementary angles?
 
  • #3
gneill said:
Have you considered that there might be two valid angles and that they might be complementary angles?

I can see how we could come to a different solution if we used the symmetry of the cosine,
\begin{equation}
cos(x) = cos (-x)
\end{equation}
Since the axis of polarization can be though of as extending onto both sides of the origin if we slapped it onto a unit circle, it doesn't really matter if we rotate the axis of polarization ##-x## degrees or ##\pi - x## degrees. However in this case it would give me something like ##150^{\circ}##, not ##50something^{\circ}##. Why would the complementary angle be a valid solution, instead of the supplementary angle?
 
  • #4
TheSodesa said:
My hunch is this is wrong, mainly because some people I talked to got a different value for θθ\theta (something like 53∘53∘53^{\circ}). My guess is that I made a mistake when I made the assumption about the power being equal to the amplitude of the electric field, but I don't really know what else I could have done there...
Yours is also right. This is due to the nature of ##y=\sin 2\theta##. If you plot this function, you will find that there are two values of ##\theta## less than 90 degrees which have the same ##y##.
 
  • #5
You have two polarizers where the second one has its orientation fixed by the given conditions to be at 90° to the polarization of the incoming light. The first polarizer will drop the intensity of the incoming light but also re-orient its polarization to match that of the angle of the polarizer. When the resulting light encounters the second polarizer the effective angle of the light with respect to the polarizer is 90° - Φ, where Φ is the rotation angle of the first polarizer with respect to the incoming light.

Consider that the power is proportional to the intensity, and the Law of Malus for intensity is

##I = I_o cos^2(\phi)##

For the two polarizers then,

##I = I_o cos^2(\phi) cos^2(90° - \phi)##

You can interchange the order of the two angles without changing the result. So there is an angle and its complement that will satisfy the equation. I'll bet the angle that your classmate found was 90 - 31.72 = 58.28 degrees.
 
  • Like
Likes blue_leaf77
  • #6
blue_leaf77 said:
Yours is also right. This is due to the nature of ##y=\sin 2\theta##. If you plot this function, you will find that there are two values of ##\theta## less than 90 degrees which have the same ##y##.

Hmm, you're right. I guess this is the result of me not really ever taking an extensive trig course and relying too much on my calculator. Since our sine has ##2\theta## instead of just ##\theta##, you can fit two angles inside ##90^{\circ}## that satisfy the equation. I keep forgetting about the other possible angles...

Thanks for the feedback.
 
  • #7
gneill said:
You have two polarizers where the second one has its orientation fixed by the given conditions to be at 90° to the polarization of the incoming light. The first polarizer will drop the intensity of the incoming light but also re-orient its polarization to match that of the angle of the polarizer. When the resulting light encounters the second polarizer the effective angle of the light with respect to the polarizer is 90° - Φ, where Φ is the rotation angle of the first polarizer with respect to the incoming light.

Consider that the power is proportional to the intensity, and the Law of Malus for intensity is

##I = I_o cos^2(\phi)##

For the two polarizers then,

##I = I_o cos^2(\phi) cos^2(90° - \phi)##

You can interchange the order of the two angles without changing the result. So there is an angle and its complement that will satisfy the equation. I'll bet the angle that your classmate found was 90 - 31.72 = 58.28 degrees.

That does sound familiar. Blue_leaf77 above also noted that my answer could be valid, although they gave a different justification for it. I do see what you mean, though, so thank you. I will mark this as solved.
 

1. How do two linear polarizers in series affect the intensity of light passing through?

When two linear polarizers are placed in series, the intensity of light passing through will decrease. This is because the polarizers are aligned in a way that allows only a certain orientation of light waves to pass through, leading to a reduction in the overall intensity.

2. Can two linear polarizers in series completely block all light?

Yes, if the two polarizers are aligned at a 90 degree angle to each other, they will completely block all light passing through. This is because the first polarizer will block all light waves with a certain orientation, and the second polarizer will block any remaining light waves that are not aligned with its orientation.

3. What is the purpose of using two linear polarizers in series?

The purpose of using two linear polarizers in series is to control the intensity and direction of light passing through. By adjusting the angle between the polarizers, the amount of light that is transmitted can be controlled. This is useful in various applications such as photography, LCD screens, and optical instruments.

4. Can two linear polarizers in series change the color of light?

No, two linear polarizers in series cannot change the color of light. They only filter out certain orientations of light waves, but do not alter the wavelength or color of the light passing through.

5. What happens when two linear polarizers with different orientations are placed in series?

If two linear polarizers with different orientations are placed in series, the intensity of light passing through will decrease, but not to the same extent as when two polarizers with the same orientation are used. This is because some light waves will still be able to pass through the second polarizer if they are aligned with its orientation.

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