# Why does light bend when it enters a medium?

I am asking this question after referring to @benjimin answers at Why does light bend?

I found the same answer at wiki. But i found a video on youtube by doctor don (https://www.youtube.com/watch?v=NLmpNM0sgYk&t=607s) claiming that method to be wrong and here is why-

According to the old explanation the waves add up and move at at different angle due to slowed speed. It looks somewhat like this - (image from the vid)

But there are also other wavefronts adding up together and going different ways.

like-

So with different wavefronts aligning in different direction, many waves propagate in different directions and this is not observed in reality and hence this explanation proves wrong. Now, Doctor Don explains that we can consider the electric field components of the waves and deuce change on them on entering a medium. Since the surface is shared by air and medium, two set of Maxwell's equation are formed and equated- (each for air and medium)

This implies that $$\frac{dB}{dt}$$ is same in both air and medium but why is this so?? Shouldn't the magnetic field also change due to wave interference with magnetic field created by oscillating electron?

• The video seems to imply that Feynman is wrong? Optics: The Principle of Least Time Commented Jan 14, 2022 at 8:25
• How do you explain a single photon bending its trajectory when it enters a medium? Commented Jan 15, 2022 at 3:49
• @Bill Alsept physics.stackexchange.com/questions/153904/…. This may help. The bending explanation is included in the comment section. Commented Jan 15, 2022 at 8:30
• @Zitscxø the question you linked seems to be about the speed of light in a medium and not about the question above Commented Jan 15, 2022 at 9:03
• @Bill Alsept i know, same question was asked by Paul in the comment section. IT was explained there. Commented Jan 15, 2022 at 11:36

Those equations are written for the $$E$$- and $$B$$-fields at the boundary. The $$E$$-field has a discontinuity at the boundary, but the $$B$$-field is continuous at the boundary, and this can be deduced by Ampere's law.

We need to make some notes first. In the video, Don Lincoln is considering the case of $$p$$-polarized light coming to the air-glass interface. In $$p$$-polarized light, the $$B$$-field is parallel to the surface of the interface.

One can use Maxwell's equations to show that the components of the $$E$$- and $$B$$-fields that are parallel to the surface must be the same in the limit $$z\rightarrow 0+$$ from above and in the limit $$z\rightarrow 0-$$ from below where $$z=0$$ is the level of the surface.

Since in the case of $$p$$-polarized light we have that $$\vec{B}(x, y, z, t)$$ is parallel to the surface at all $$(x, y, z, t)$$, our proof will demonstrate that $$\vec{B}_{\text{air}}(z\rightarrow 0+) = \vec{B}_{\text{glass}}(z\rightarrow 0-)$$.

To show our claim, we will use Ampere's law. At the air-glass interface, consider a rectangular loop $$C$$ of height $$h$$ and length $$\ell$$ half-way between the the two media as depicted below.

By Ampere's law, we have $$\oint_{C} \vec{B}\cdot \text d\vec{l} = \iint_{S} \mu \left[ \vec{J} + \epsilon\frac{\partial\vec{E}}{\partial t} \right] \cdot \text d\vec{A}$$ where the right-hand side is an integral over the rectangle whose boundary is loop $$C$$. The absolute value of the area integral on the right-hand side is bounded above by $$\text{|maximum of integrand|}\times \text{Area}$$ where $$\text{Area} = h\ell$$. As we send $$h\rightarrow 0$$, we see that $$\text{Area} = h\ell\rightarrow 0$$, so the area integral in the above equation goes to zero. Hence in the limit of $$h\rightarrow 0$$, we have $$\oint_{C} \vec{B}\cdot \text d\vec{l}\rightarrow 0.$$ Thus, $$\vec{B}(z\rightarrow 0+)\cdot \text d\vec{l}_{1} + \vec{B}(z\rightarrow 0-)\cdot \text d\vec{l}_{2} = 0$$ for any length element $$\text d\vec{l}_{1}$$ parallel to the surface. Since the loop goes in opposite directions above and below the surface, $$\text d\vec{l}_{1} = -\text d\vec{l}_{2}$$ and thus $$\vec{B}(z\rightarrow 0+)\cdot \text d\vec{l}_{1} + \vec{B}(z\rightarrow 0-)\cdot (-\text d\vec{l}_{1}) = 0$$. Since we can orient the length element $$\text d\vec{l}_{1}$$ any way we want as long as it is parallel to the interface surface and since $$\vec{B}$$ is parallel to the interface surface in $$p$$-polarized light, it follows that $$\vec{B}(z\rightarrow 0+) - \vec{B}(z\rightarrow 0-) = 0$$, and so $$\vec{B}(z\rightarrow 0+) = \vec{B}(z\rightarrow 0-)$$, as desired.

Thus, $$\partial\vec{B}/\partial t$$ is the same as $$z\rightarrow 0+$$ and $$z\rightarrow 0-$$. The behavior of $$\vec{B}$$ does change as you pass from air to glass, but we've shown that it is same at the boundary, meaning it is continuous as you pass through the boundary. This is exactly the same way the parallel component $$\vec{E}_{\text{parallel}}$$ of $$\vec{E}$$ is same at the boundary and continuous across the boundary, as the video claims. In fact, a similar proof of this can be devised using Faraday's law.

(Lastly, I should repeat that the $$B$$-field is parallel to the interface surface only in the case of $$p$$-polarized light. In the case of $$s$$-polarized light, things are a bit different. The same reasoning above applies, but it only applies to the component of $$\vec{B}$$ that is parallel to the interface surface.)

• Thank you for clarifying. The term 'boundary condition' cleared my misunderstanding. The proof was insightful. Commented Jan 14, 2022 at 9:14
• Tho your reply fast. I didn't expect a answer this fast. Commented Jan 14, 2022 at 9:19