<h4>Statement</h4>
<p>
$13.2.11$
What minimum angle of incidence should have a light ray that incides over a group of plane transparent plates (each one with refraction index decreasing $k$ times respect to the upper to it) such that the ray doesn't pass the group? The refraction index of the upper plate is $n$ and the amount of plates is $N$.
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" alt="13.2.11" width="300" />
<figcaption>
For problem 13.2.11
</figcaption>
</figure>
</center>
<h4>Solution</h4>
<p class="TxtSolutions">
Let's consider the following figure
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="13.2.11" width="450" />
<figcaption>
Ray's path through plates
</figcaption>
</figure>
</center>
<p class ="TxtSolutions">
Applying Snell's law for each interphase between media
</p>
<center>
$n_0 \sin{\alpha} = n_1 \sin{\alpha_1} = n \sin{\alpha_1}$
</center>
<center>
$n \sin{\alpha_1} = n_2 \sin{\alpha_2}$
</center>
<center>
$n_2 \sin{\alpha_2} = n_3 \sin{\alpha_3}$
</center>
<center>
$\vdots$
</center><center>
$n_{N-1} \sin{\alpha_{N-1}} = n_N \sin{\frac{\pi}{2}} = n_N$
</center>
<p calss="TxtSolutions">
It's known that $n_2 = \frac{n}{k}$, $n_3 = \frac{n_2}{k}=\frac{n}{k^2}$, and so on. Then, $n_N=\frac{n}{k^{N-1}}$. The $N$-th refraction angle is $\frac{\pi}{2}$ beacause the ray doesn't pass this latest plate, i.e., there is a total reflection.
</p>
<center>
$n_m \sin{\alpha} = \frac{n}{k^{N-1}}$
</center>
<p class="TxtSolutions">
If exterior medium is air, $n_m = 1$,
</p>
<center>
$\boxed{\sin{\alpha}=\frac{n}{k^{N-1}}}$
</center>
<p class="TxtSolutions">
This angle $\alpha$ is the critical angle of the internal total reflection for this group of plates.
</p>
<p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
</p>
</article>