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Light is an electromagnetic wave composed of electric and magnetic components.

I recently read that the velocity of light is in the direction $\mathbf{E}\times\mathbf{B}$ where $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic field vectors.

Now, suppose at some point in space, the electric and magnetic fields are $\mathbf{E}$ and $\mathbf{B}$, which are both functions of time, suppose that we place a mirror there, perpendicular to the incident light, then, due to reflection from denser medium, there will be a phase difference of $\pi$ in the electric and magnetic field components of the wave. So, the electric and magnetic fields are $-\mathbf{E}, -\mathbf{B}$ after reflection. Hence, the direction of propagation of light after reflection is the same as the direction of $(\mathbf{-E})\times(\mathbf{-B})=\mathbf{E}\times\mathbf{B}$ which means that the light wave is still traveling in the same direction.

Where is the error in this?

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From Wikipedia, Reflection Phase Change

"Phase" here is the phase of the electric field oscillations, not the magnetic field oscillations (while the electric field will undergo 180° phase change, the magnetic field will undergo 0° phase change.

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Suppose interface to be $xy$-plane then $$\mathbf{E}_I(z,t)=E_{0I} \ e^{i(k_1z-\omega t)}\hat{x}$$ $$\mathbf{B}_I(z,t)=\frac{1}{v_1}E_{0I} \ e^{i(k_1z-\omega t)}\hat{y}$$

It gives rise to a reflected wave, $$\mathbf{E}_R(z,t)=E_{0R} \ e^{i(-k_1z-\omega t)}\hat{x}$$ $$\mathbf{B}_R(z,t)=-\frac{1}{v_1}E_{0R} \ e^{i(-k_1z-\omega t)}\hat{y}$$ Now if $v_2<v_1$ then there will be phase difference introducing this to reflected waves $$\mathbf{E}_R(z,t)=E_{0R} \ e^{i(-k_1z-\omega t+\pi)}\hat{x}$$ $$\mathbf{B}_R(z,t)=-\frac{1}{v_1}E_{0R} \ e^{i(-k_1z-\omega t+\pi)}\hat{y}$$

$$\mathbf{E}_R(z,t)=-E_{0R} \ e^{i(-k_1z-\omega t)}\hat{x}$$ $$\mathbf{B}_R(z,t)=\frac{1}{v_1}E_{0R} \ e^{i(-k_1z-\omega t)}\hat{y}$$

The direction of wave propagation: $-\hat{x}\times\hat{j}=-\hat{k}$

It should be.


Reference : Sec. 9.3.2 Electrodynamics : DJ Griffith

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  • $\begingroup$ Also, can we explain it using the fact that magnetic field is a pseudo vector? $\endgroup$
    – mathsworm
    Sep 28 at 9:29

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