The reflection at a dielectric interface was analysed in Griffiths introduction to electrodynamics using the following diagram. enter image description here

I do not understand why the direction of $\vec E_r$ and $\vec B_r$ are as given. Why can't it be as drawn in a different colour?

I calculated with the new $\vec E_r$ and $\vec B_r$ and got a different result - with $\alpha$ and $\beta$ interchanged, implying that $180^o$ phase shift happens in reflection from rarer medium.

I need to know why at the interface, the reflected $\vec E_r$ and $\vec B_r$ should be as drawn by griffith.

Here are my calculation : enter image description here

The equations of griffiths were:

$E_{0i} + E_{0r} =\alpha E_{0t} $

$E_{0i} - E_{0r} =\beta E_{0t} $


1 Answer 1


Yes, you got a different result because you started by defining the direction of a positive field differently.

If you reverse the defined directions of the reflected E-field and B-field (which you are free to do), then don't be surprised if your reflection coefficient reverses sign.

This is perhaps more obvious to see at normal incidence, where the reflection coefficients are either $$ r = \pm \frac{n_2-n_1}{n_1+n_2}$$ depending on whether you started by defining the reflected electric field in the opposite or the same direction as the incident electric field.

  • $\begingroup$ But both results are different right... with $+$ sign, it means that reflected wave is out of phase with incident wave when $n_2 < n_1$. ie : reflection at rarer medium. But with $-$ sign, this happens for $n_2>n_1$. ie: reflection at denser medium. We know that only one is true. ( with $-$ sign) $\endgroup$ Oct 28, 2020 at 8:09
  • $\begingroup$ @RishabNavaneet No. If you have $n_2>n_1$ and start by drawing the incident and reflected electric fields in the same direction, then $r$ will be negative. If you start by drawing them in opposite directions then $r$ will be positive. Only in the first case can the sign of $r$ be directly interpreted as a phase change. If you do the derivation with a transmitted E drawn in the opposite direction to incident E you would get a negative transmission coefficient, but this is not a phase change... $\endgroup$
    – ProfRob
    Oct 28, 2020 at 8:17
  • $\begingroup$ Yeahh... I got it now. The electric Field which I drew is already out of phase with incident wave. The reflection of a reflection should give us back the same thing. If I reflect the reflected wave about xy plane again, I see that my arrow points opposite to incident $\vec E$. So it was just the issue with positive direction. Thankyou.. $\endgroup$ Oct 28, 2020 at 8:30

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