# Light entering in a 0° angle, is there any reflection?

Is there any reflection of light that enters a new medium at a 0° angle, if the electric field is such, that it is completely in the plane of the double layer?

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How would you propose to conserve momentum in such a scenario? – Kyle Oman Aug 1 '13 at 14:46
okay, so no light is reflected right? – Xin Wang Aug 1 '13 at 15:00
I don't actually see where the momentum argument is getting you. If light was reflected of the body, momentum would have to be transferred onto the body to compensate but that's not uncommon with reflection. – Jonas Aug 1 '13 at 15:50
Hm I think I had the situation rotated 90 degrees in my head, and the momentum argument was a bit shaky besides... – Kyle Oman Aug 2 '13 at 17:37

Now, according to Fresnel equations, we have: $$\tilde E_{0R}=\left(\frac{\alpha -\beta }{\alpha +\beta}\right)\tilde E_{0I} \tag{1}$$ $$\tilde E_{0T}=\left(\frac{2 }{\alpha +\beta}\right)\tilde E_{0I}\tag{2}$$ where $\alpha=\dfrac{\cos \theta_T}{\cos \theta_I}$ and $\beta=\dfrac{\mu_1 n_2}{\mu_2 n_1}$.
Now, apply Snell's law to find the relation between $\theta_T$ and $\theta_I$: $$\frac{\sin \theta_T}{\sin \theta_I}=\frac{n_1}{n_2}\tag{3}$$
In the case of zero incidence angle ($\theta_I=0\text{ or }\cos\theta_I=1$) , $(3)$ implies that $\cos \theta_T=1$ and according to $(1)$, it is evident that $\tilde E_{0R}\neq 0$: $$\alpha=1\to \cases{\tilde E_{0R}=\left(\frac{1-\beta }{1 +\beta}\right)\tilde E_{0I}\\\tilde E_{0T}=\left(\frac{2 }{1 +\beta}\right)\tilde E_{0I}}$$