<h3 id="back-link"><a href="/">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <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>

    <h3>Solution</h3>
    <p>
      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>
      Applying Snell's law for each interphase between media
      $$n_0 \sin{\alpha} = n_1 \sin{\alpha_1} = n \sin{\alpha_1}$$
      $$n \sin{\alpha_1} = n_2 \sin{\alpha_2}$$
      $$n_2 \sin{\alpha_2} = n_3 \sin{\alpha_3}$$
      $$\vdots$$
      $$n_{N-1} \sin{\alpha_{N-1}} = n_N \sin{\frac{\pi}{2}} = n_N$$
      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.
      $$n_m \sin{\alpha} = \frac{n}{k^{N-1}}$$
      If exterior medium is air, $n_m = 1$,
      $$\boxed{\sin{\alpha}=\frac{n}{k^{N-1}}}$$
      This angle $\alpha$ is the critical angle of the internal total reflection for this group of plates.
    </p>


    <p 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>



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