# Condition of continuity of E field at a boundary

I am trying to understand the derivation of Snell's law using Maxwell's equation and got stuck.

My text book says that "the E field that is tangent to the interface must be continuous" in order to consider refraction of light.
If it were static E field I understand this is true because in electrostatics

rotE = 0

holds. However Snell's law describes how electromagnetic waves change their direction of propagation when going through an interface of two mediums. Since our E filed is changing dynamically, we should use the equation

rotE = -∂B/∂t

in stead. To me it is not obvious why this equation leads to the continuity condition.
How does the continuity condition in Snell's law appears from Maxwell's equations?

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Meir Achuz
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The continuity of E tangential comes from applying Stokes' theorem to rotE = -∂B/∂t.
The area for $$\int{\bf dS}\partial_t{\bf B}$$ shrinks to zero.

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The continuity of E tangential comes from applying Stokes' theorem to rotE = -∂B/∂t.
The area for $$\int{\bf dS}\partial_t{\bf B}$$ shrinks to zero.
Stokes's theorem for rotE is
$$\int{ rot\bf{E}}{\bf dS}= \oint _{∂S} \bf Edx = \oint _{∂S} \bf \partial_t B dx$$

How does this lead to the continuity condition?

OK I see. It seems like the continuity condition is something to do with the fact that the interface has zero volume and the planar surface is sufficiently large.
The path of line integration must be an infinitely thin rectangular when the area of the box approaches to 0.
The shape of the box is the key because it will allow the line integration to become 0 even if rotE is non-zero.
Thanks for all the replies.

Will the same rule apply if there is a gradient layer between the two phases?
Let's say the geometry is no longer flat but curved, and the curvature of radius is comparable to the thickness of gradient layer.
I'm pretty sure that the parallel component of the E field strength will still be continuous at any point.
But will the phase still be the same?

Simon Bridge