# Using Snell's law in transmission and reflection coefficients

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

$$r=\frac{n_2\frac{\cos\theta_i\sin\theta_t}{\sin\theta_i}-n_2\cos\theta_t}{n_2\frac{\cos\theta_i\sin\theta_t}{\sin\theta_i}+n_2\cos\theta_t}=\frac{\cos\theta_i\sin\theta_t-\cos\theta_t\sin\theta_i}{\cos\theta_i\sin\theta_t+\cos\theta_t\sin\theta_i}=-\frac{\sin\left(\theta_i-\theta_t\right)}{\sin\left(\theta_i+\theta_t\right)}.\tag{2}$$

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?

• @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. Aug 28, 2021 at 8:35
• Check the derivation of eq. 2, can you really substitute 0? Aug 28, 2021 at 8:36
• 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. Aug 28, 2021 at 8:59

## 1 Answer

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

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