Phase Change and Reflection of Electromagnetic Waves

[Abhishek kumar]

Isn't "t" same in the two wave expressions ?

Aren't we considering the same instant ?

Read standing wave form by super position of reflected and incident wave

 

[Delta2]

But presence of minus sign in reflected wave doesn't mean complete cancellation.

E₀cos (kz - ωt) - E₀cos (kz + ωt) ≠ 0

They completely cancel at z=0, if reflection happens at z=0. If it happens at z=a you need the phase term $2ka$ as Charles says at post #2.

 

[Abhishek kumar]

They completely cancel at z=0, if reflection happens at z=0. If it happens at z=a you need the phase term $2ka$ as Charles says at post #2.

That's what i want to say

 

[Jahnavi]

You are only considering position not time.when incident wave traveled a unit reflected wave just started

That's what i want to say

This is not what you said in post 27

 

[Abhishek kumar]

This is not what you said in post 27

You have taken same distance traveled by both the wave at any instant of time is it possible?that's why i said you r considering only same position.For same time position will be different

 

[Charles Link]

At z = a , E₀cos (ka- ωt) - E₀cos (ka + ωt) = 2E₀sin(ka)sin(ωt) ≠ 0

Can you show how they cancel ?

@Jahnavi I think you have this part figured out already, but to show it in detail: $\\$ $E_{incident}(z,t)=E_o \hat{i} \cos(kz-\omega t)$, and $\\$ $E_{reflected}(z,t)=-E_o \hat{i} \cos(2ka-kz-\omega t)$. $\\$ Evaluating at $z=a$: $\\$ $E_{incident}(a,t)=E_o \hat{i} \cos(ka-\omega t)$, and $\\$ $E_{reflected}(a,t)=-E_o \hat{i} \cos(2ka-\omega t-ka)=-E_o \hat{i} \cos(ka-\omega t)$. $\\$ Thereby, they cancel at $z=a$. $\\$ Also, in post 21, you asked about the vector $\hat{i}$. Yes, that means the electric field for this electromagnetic wave is transverse and points in the x-direction.