- #1

- 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...