# E and B Fields of Monochromatic Plane Waves

• roam
In summary, the light is polarized parallel to the ##(y-z)## plane. It is collimated and passes as a parallel beam in a direction perpendicular to the ##x+y+z=0## plane. It is intensity is ##1 \ mW/cm^2##.
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.

roam said:
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##.

blue_leaf77 said:
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}## ?

roam said:
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}##.

Last edited:
blue_leaf77 said:
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?

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.

## 1. What are E and B fields of monochromatic plane waves?

E and B fields are two components of electromagnetic waves, which are transverse in nature. They are perpendicular to each other and to the direction of propagation. E-field represents the electric field and B-field represents the magnetic field.

## 2. How are E and B fields related in monochromatic plane waves?

In monochromatic plane waves, the E and B fields are related by the speed of light, c. The ratio of E-field to B-field is equal to the speed of light in a vacuum, which is approximately 3 x 10^8 meters per second.

## 3. What is the significance of a monochromatic plane wave?

A monochromatic plane wave is a simplified model used to study the behavior of electromagnetic waves. It helps in understanding the fundamental properties of light, such as its wavelength, frequency, and speed.

## 4. How are the E and B fields affected by the medium through which the monochromatic plane wave travels?

The E and B fields of a monochromatic plane wave are affected by the properties of the medium through which it travels. These properties include the refractive index, conductivity, and permeability of the medium. These factors can change the direction, speed, and amplitude of the wave.

## 5. Can E and B fields of monochromatic plane waves be measured?

Yes, E and B fields of monochromatic plane waves can be measured using specialized instruments such as electromagnetic field probes. These instruments can detect the strength and direction of the fields and convert them into measurable quantities.

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