E and B Fields of Monochromatic Plane Waves

[roam]

Homework Statement

560 nm light is collimated and passes as a parallel beam in a direction perpendicular to the $x+y+z=0$ plane. It is polarized parallel to the $(y-z)$ plane. Treating it as a plane wave, what are the real E and B fields?

Intensity of the beam is $1 \ mW/cm^2$. Make your expressions for the fields as numerical as possible.

Homework Equations

$E(r, t) = E_0 \ cos (k.r - \omega t) \hat{n}$

$B(r,t) = \frac{1}{c} E_0 \ cos (k.r - \omega t) (\hat{k} \times \hat{n})$

Where $k$ is the propagation/wave vector, and $\hat{n}$ is the direction of popularization.

$k=\frac{\omega}{\lambda f}=\frac{\omega}{c}= \frac{2 \pi}{\lambda}$

Using the following relationship, I used the given intensity to work out the amplitude:

$I=\frac{1}{2} c \epsilon_0 E_0^2 \implies E_0 = \sqrt{\frac{2I}{c \epsilon_0}} = 7.685 \times 10^{-10} \ V/m$

The Attempt at a Solution

So, to find $k.r$ and $k \times \hat{n}$, I need the equation for a plane perpendicular to the $x+y+z=0$ plane. But how can I find this equation when there are infinitely many such vectors?

For instance the wave vector is $k= \omega/c \ \hat{k}$, where $\hat{k}$ is a unit vector in the direction of the vector perpendicular to the $x+y+z=0$ plane. I also need that in order to work out the cross product $\hat{k} \times \hat{n}$ for the B field.

As for the polarization $\hat{n}$ (parallel to y-z plane), would the unit vector be $\hat{n} = \frac{\hat{y}+\hat{z}}{\sqrt{2}}$? Or does the plus sign need to be changed to a minus? Any explanation would be greatly appreciated.

 

[blue_leaf77]

But how can I find this equation when there are infinitely many such vectors?

No, there is only one unit vector perpendicular to a given plane. Given $\lambda$ and the unit vector associated with the propagation, $k$ will be uniquely defined.
After finding such unit vector, use the fact that $k$ must be perpendicular to $n$.

 

[roam]

No, there is only one unit vector perpendicular to a given plane. Given $\lambda$ and the unit vector associated with the propagation, $k$ will be uniquely defined.
After finding such unit vector, use the fact that $k$ must be perpendicular to $n$.

Yes, we need to have $\hat{n} . k = 0$. But what is the actual expression for this unit vector (that is perpendicular to $x+y+z=0$)?

(I believe it should be of the form $\hat{k}=(a\hat{x}+b\hat{y}+c\hat{z})/\sqrt{3}$, I then need to write $k=\frac{2 \pi}{\lambda} \hat{k}$).

And what expression do you use for $\hat{n}$? Is it $(\hat{y} + \hat{z})/\sqrt{2}$ or $(\hat{y} - \hat{z})/\sqrt{2}$ ?

 

[blue_leaf77]

Yes, we need to have $\hat{n} . k = 0$. But what is the actual expression for this unit vector (that is perpendicular to $x+y+z=0$)?

Note that you can view the expression $x+y+z=0$ as a dot product between two vectors which are perpendicular to each other.
Another alternative will be apparent if you are more familiar with the use of gradient $\nabla$ to find a normal vector w.r.t. to a surface.
Let's first solve the vector $k$, after that it should be easy to find the correct $\hat{n}$.

 

[roam]

Note that you can view the expression $x+y+z=0$ as a dot product between two vectors which are perpendicular to each other.
Another alternative will be apparent if you are more familiar with the use of gradient $\nabla$ to find a normal vector w.r.t. to a surface.
Let's first solve the vector $k$, after that it should be easy to find the correct $\hat{n}$.

How should I use the gradient $\nabla$ here? What are we differentiating?

 

[blue_leaf77]

It depends on which one is easier to you for use, if you think you can find the required unit vector for $k$ by viewing $x+y+z=0$ as a dot product as I said above then take this way.
As for the gradient, read this http://mathworld.wolfram.com/NormalVector.html.