If we have an electromagnetic wave propagating between two media with refractive indices $n_1$ and $n_2$. The reflection and transmission coefficients at the interface can be written from `Optics, Light, and Lasers' by Dieter Meschede, as

$$r=\frac{E_{r}}{E_i}=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\qquad\qquad t=\frac{E_{t}}{E_i}=\frac{2n_1\cos\theta_i}{n_1\cos\theta_i+n_2\cos\theta_t},\tag{1}$$

where the indices $i$, $r$, and $t$ stand for incident, reflected and transmitted, $E$ is the electric field, and $\theta$ the angle made between the direction of propagation and the normal to the interface such that, Snell's law is written $n_1\sin\theta_i=n_2\sin\theta_t$. Using Snell's law and substituting into Eq.(1) for $n_1$, the reflection and transmission coefficients can be written as


and for $t$ $$t=\frac{2\cos\left(\theta_i\right)\sin\left(\theta_t\right)}{\sin\left(\theta_i+\theta_t\right)}\tag{3}.$$

Written in the form of Eq.(2 & 3) seems to involve an inconsistency not observed in Eq.(1). If the angle of incidence $\theta_i=0$ then Snell's law tells us that the transmitted angle will be $\theta_t=0$, using Eq.(1) we have

$$r=\frac{n_1-n_2}{n_1+n_2},\qquad\qquad t=\frac{2n_1}{n_1+n_2}.\tag{4}$$

However, using Eq.(2 & 3) we have

$$r=-\frac{\sin(0)}{\sin(0)},\qquad\qquad t=2\frac{\sin(0)}{\sin(0)}.\tag{5}$$

Eq.(4) and Eq.(5) are clearly different, one is finite and gives a sensible result, whilst the other is not defined. There seems to be an error in my reasoning?

  • $\begingroup$ @OfekGillon Substituting Snell's law into Eq1 for n1. Eq2 can also be found in optics books such as `Optics, Light, and Lasers' by Dieter Meschede. $\endgroup$
    – jamie1989
    Aug 28, 2021 at 8:35
  • $\begingroup$ Check the derivation of eq. 2, can you really substitute 0? $\endgroup$ Aug 28, 2021 at 8:36
  • $\begingroup$ Eq2 & 3 seem fine to me. I suppose that is my question, putting 0 for the angles is fine in one but not the other. $\endgroup$
    – jamie1989
    Aug 28, 2021 at 8:59

1 Answer 1


The answer is that $\frac{0}{0}$ is just not well defined. To get eq. 2, you need to multiply both the numerator and denominator in $\sin \theta_t $ which is $0$ for $\theta_i = 0 $, meaning this is not mathematically allowed for the case you wrote. However, you can check what it's approaching for $\theta_i \to 0$ and by taking L'hospital's rule and deriving both parts of the fraction with respect to $\theta_i$, keeping in mind that $\frac{d\theta_t}{d\theta_i} |_{\theta_i=0} = \frac{n_1}{n_2} $, you'll get exactly eq 4

  • $\begingroup$ I'm not sure I understand, what equation are you multiplying the numerator and denominator by $\sin\theta_t$? $\endgroup$
    – jamie1989
    Aug 28, 2021 at 10:45
  • $\begingroup$ As you wrote in your derivation there is a part you multiplies and divided something by $\sin \theta_t $ right? $\endgroup$ Aug 28, 2021 at 10:55

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