Subfield: Optics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Consider a uniform isosceles triangle prism $\mathrm{ABC}$, with the apex angle $\theta=110^{\circ}$ at vertex $\mathrm{A}$. One of the sides, $\mathrm{AC}$, is coated with silver, allowing it to function as a mirror. When a monochrome light-ray of wavelength $\lambda$ approaches side $\mathrm{AB}$ at an angle of incidence $\alpha$, it first refracts, then reaches side AC, reflects, and continues to base BC. After another refraction, the ray eventually exits the prism at the angle of emergence which is also equal to the angle of incidence (see Fig. A).
<img_4302>



What is the relative refractive index of the prism for that particular wavelength $\lambda$ with respect to the outside environment, given that $\alpha=70^{\circ}$.

Solution:
The light-path refracts on side $A B$ at point $M$, reflects on side $A C$ at point $N$ and refracts on base $\mathrm{BC}$ at point $\mathrm{P}$ (see Fig. A). Define the angle of refraction inside the prism to be $\beta$, then from Snell's law:
$$
\sin \alpha=n \sin \beta .
\tag{29}
$$

From the law of reflection and the $180^{\circ}$-sum of three interior angles inside any triangles:
$$
\begin{aligned}
\widehat{\mathrm{MNA}} & =180^{\circ}-\widehat{\mathrm{NAM}}-\widehat{\mathrm{AMN}}=180^{\circ}-\theta-\left(90^{\circ}-\beta\right) \\
& =\widehat{\mathrm{PNC}}=180^{\circ}-\widehat{\mathrm{NCP}}-\widehat{\mathrm{CPN}}=180^{\circ}-\left(\frac{180^{\circ}-\theta}{2}\right)-\left(90^{\circ}+\beta\right),
\end{aligned}
$$
we obtain the refraction angle $\beta$ to be:
$$
\beta=\frac{3 \theta-180^{\circ}}{4}
$$
Plug this finding into Eq. (29), we get the relative refraction index of the prism with respect to the outside environment:
$$
n=\frac{\sin \alpha}{\sin \beta}=\left.\frac{\sin \alpha}{\sin \left(\frac{3 \theta-180^{\circ}}{4}\right)}\right|_{\alpha=70^{\circ}, \theta=110^{\circ}} \approx 1.5436
$$

Final answer: 1.5436