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The intensity of an electromagnetic wave is defined as the magnitude of the time-average of the Poynting vector:

$$ I = \left|\left<\vec{S}\right>_t\right| $$

where $\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$.

For a plane wave $\vec{E} = {E}_0 e^{i(\vec{k} \cdot \vec{r}-\omega t)} $ one obtains

$$I = \frac{1}{2}c\epsilon_0 |\vec{E}_0|^2 $$

using $\vec{B}_0 = \vec{k} \times \vec{E}_0$, which follows from Maxwells equations.

From every slit, a spherical wave emerges (Huygens principle):

$$\vec{E} = \frac{\vec{E}_0}{r} e^{i(kr-\omega t)} $$ However in the context of the double slit experiment (or any diffraction/interference experiment) the intensity is given as

$$ I \propto | \vec{E} |^2 $$

in various textbooks and articles.

I do not see how this expression of $I$ is consistent with the definition above.

Thank you for your help!

Edit

Take a look at this similar post: Exact relationship between electric field and intensity

In the first answer, the person writes in the second line

"$ |\vec{H}| = \sqrt{\frac{\epsilon_0}{\mu_0}} |\vec{E}|$ for an electromagnetic wave in vacuum"

But this relation (as far as I know) only holds for plane waves, right? And very unfortunately, plane waves do not emerge from the double slits.

OkTennis
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2 Answers2

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Let's find out the expression for the magnetic field for a spherical wave, considering $\mathbf E = \mathbf{\hat{\theta}}\frac{E_0}{r} e^{i(kr-\omega t)}$ (the wave is travelling in the radial direction, so I choosed the system $(\hat{r},\hat{\theta},\hat{\phi})$ conveniently). What we should do is to consider Faraday's law

$$ \nabla \times \mathbf E = - \frac{\partial \mathbf B}{\partial t}. $$

In spherical coordinates,

$$ \nabla \times \mathbf E = \frac{1}{r}\frac{\partial }{\partial r}( rE_{\theta})\hat{\phi} = \hat{\phi} ik\frac{E_0}{r}e^{i(kr-\omega t)} \equiv i \mathbf k \times \mathbf E, $$

since $\mathbf k = k \hat{r}$. By the first equation

$$ \mathbf B(r,t) = - \int^t (\nabla \times \mathbf E(r,t')) dt' = \hat{\phi} k \frac{E_0}{\omega r} e^{i(kr-\omega t)} = \frac{\hat{r}}{c}\times \mathbf E. $$

You see that $\mathbf B \perp \mathbf{E}$, and $E = cB$, and the expression follows.

Ruffolo
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You are correct:

The exact calculation of the power density has 1/2 the power in E and 1/2 the power in B. In a simple plane wave in vacuum, it is safe to ignore B and double the value computed from E.

There is no need to drag B around, so basically: it's a shorthand that gives the same result. We're here to focus on the interference.

Now if you have non-linear and/or non-isotropic materials, you may need to be more cautious.

JEB
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