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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.