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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| $13.2.11$ | | $13.2.11$ |
| 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 $N$. | | 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> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="13.2.11" width="300" /> | | loading="lazy" alt="13.2.11" width="300" /> |
| <figcaption> | | <figcaption> |
| For problem 13.2.11 | | For problem 13.2.11 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider the following figure | | Let's consider the following figure |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" alt="13.2.11" width="450" /> | | loading="lazy" alt="13.2.11" width="450" /> |
| <figcaption> | | <figcaption> |
| Ray's path through plates | | Ray's path through plates |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Applying Snell's law for each interphase between media | | Applying Snell's law for each interphase between media |
| $$n_0 \sin{\alpha} = n_1 \sin{\alpha_1} = n \sin{\alpha_1}$$ | | $$n_0 \sin{\alpha} = n_1 \sin{\alpha_1} = n \sin{\alpha_1}$$ |
| $$n \sin{\alpha_1} = n_2 \sin{\alpha_2}$$ | | $$n \sin{\alpha_1} = n_2 \sin{\alpha_2}$$ |
| $$n_2 \sin{\alpha_2} = n_3 \sin{\alpha_3}$$ | | $$n_2 \sin{\alpha_2} = n_3 \sin{\alpha_3}$$ |
| $$\vdots$$ | | $$\vdots$$ |
| $$n_{N-1} \sin{\alpha_{N-1}} = n_N \sin{\frac{\pi}{2}} = n_N$$ | | $$n_{N-1} \sin{\alpha_{N-1}} = n_N \sin{\frac{\pi}{2}} = n_N$$ |
| 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. | | 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. |
| $$n_m \sin{\alpha} = \frac{n}{k^{N-1}}$$ | | $$n_m \sin{\alpha} = \frac{n}{k^{N-1}}$$ |
| If exterior medium is air, $n_m = 1$, | | If exterior medium is air, $n_m = 1$, |
| $$\boxed{\sin{\alpha}=\frac{n}{k^{N-1}}}$$ | | $$\boxed{\sin{\alpha}=\frac{n}{k^{N-1}}}$$ |
| This angle $\alpha$ is the critical angle of the internal total reflection for this group of plates. | | This angle $\alpha$ is the critical angle of the internal total reflection for this group of plates. |
| </p> | | </p> |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
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