How Do You Find the Frequency of a 550 nm Harmonic Wave in Vacuum?

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Homework Help Overview

The discussion revolves around finding the frequency of a 550 nm harmonic wave traveling in vacuum, with specific focus on its electric field in the z direction and propagation in the y direction. The problem includes multiple parts, such as determining angular frequency and wave number, as well as expressions for electric and magnetic fields.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve for the amplitude of the electric field and expresses confusion regarding the constant C in their wave equation. Some participants question the setup of the electric field and its direction based on the problem statement.

Discussion Status

Participants are actively engaging in clarifying the correct form of the wave equation and the implications of the wave's direction and polarization. Guidance has been offered regarding the general form of a plane wave, prompting further exploration of the original poster's approach.

Contextual Notes

There is a noted discrepancy in the direction of the electric field and the propagation direction, which some participants are addressing. The original poster has indicated a formatting error in their expression, which may affect the interpretation of their attempts.

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Homework Statement


A 550 nm harmonic wave whose electric field is in the z direction is traveling in the y direction in vacuum.

a)find the frequency of the wave
b)determine [tex]\omega[/tex] and k.
c)if the electric field amplitude is 600 V/m find the amplitude of the magnetic field.
d)write an expression for both E(t) and B(t) given that each is zero at x=0 and t=0.


Homework Equations


[tex]\Psi[/tex](x,t)=Acos(kx-[tex]\omega[/tex]t)+Csin(kx-[tex]\omega[/tex]t)


The Attempt at a Solution


I've solved all the other questions but part d is giving me a bit of trouble.

[tex]\Psi[/tex](x,t)=Acos(kx-[tex]\omega[/tex]t)+Csin(kx-[tex]\omega[/tex]t)
[tex]\Psi[/tex](0,0)=Acos(0) + Csin(0)
0=A(1) + C(0)
therefore A=0 but not necessarily C so i wrote
E(t)=[tex]\hat{y}[/tex]Csin(kx-[tex]\omega[/tex]t) and
B(t)=[tex]\hat{z}[/tex]Csin(kx-[tex]\omega[/tex]t)

This is the part I'm stuck on, I'm unsure on how to solve for the constant C, the amplitude of the function.

*edit the formatting is a bit off, it should be kx-wt, the w isn't a power of anything.
 
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First, [itex]\textbf{E}(\textbf{r},t)=C_E\sin(kx-\omega t)\mathbf{\hat{y}}[/itex] describes an electric field, with amplitude [itex]C_E[/itex], polarized in the [itex]y[/itex]-direction, and traveling in the [itex]x[/itex]-direction...that isn't what the problem statement you've posted asks you to describe...
 
Thank you for the quick reply. Since my idea is wrong, could you give me a hint on the correct way to solve the problem.
 
Remember, the general form of a plane wave, traveling along the direction of [itex]\mathbf{k}[/itex] and polarized in the [itex]\mathbf{\hat{n}}[/itex] direction is [itex]\textbf{E}(\textbf{r},t)=C\sin(\mathbf{k}\cdot\textbf{r}-\omega t)\mathbf{\hat{n}}+D\cos(\mathbf{k}\cdot\textbf{r}-\omega t)\mathbf{\hat{n}}[/tex]<br /> <br /> What would you expect [itex]\mathbf{\hat{n}}[/itex] to be if the field is polarized in the [itex]z[/itex]-direction?<br /> <br /> What would you expect [itex]\mathbf{k}\cdot\textbf{r}[/itex] to be if the field is traveling in the [itex]y[/itex]-direction?[/itex]
 
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