IPhO 2019, Експеримент — Задача 1.3. Part C: The refractive index of a triangular prism

• C.1 $n = 2\sin\left(\frac{\langle\delta_{\min}\rangle}{2} + 30^{\circ}\right)$

• C.2 1.6433

  n = 1.6433 ± 0.0006; the literature value is 1.6425 ± 0.0007.

Section C – the refraction index of a triangular prism

C.1: From the sketch showing the path of the laser beam and from the principle the beam path reversal we get that the deflection angle $\delta$ from the direction of in the incoming beam will not change if we switch the angles $\alpha_1$ and $\alpha_2$. Thus we get that $\delta$ achieves an extremum value (in fact, a minimal value) when the situation is perfectly symmetric, that is when $\alpha_1 = \alpha_2$. In this case,

$\beta_1 = \beta_2 = \frac{\phi}{2}$.

For the symmetric case, the incident angle $\alpha$ holds the relation $\alpha = \frac{\delta}{2} + \frac{\phi}{2}$ and from Snell’s law we get $\sin\left(\frac{\delta}{2} + \frac{\phi}{2}\right) = n\sin\frac{\phi}{2}$.

If the prism is not exactly equilateral, we will mark the angles of the prism by $\phi_i = 60^{\circ} + 2\varepsilon_i$. From the sum of angles in a triangle we get $\sum\varepsilon_i = 0$. Additionally $\beta_i = 30 + \varepsilon_i$. In this case $\delta_{\min} = \delta_0 + 2\Delta_i$ where $\delta_0$ is the minimal $\delta$ when $\phi = 60^{\circ}$.

From Snell’s law we get $\sin\left(\frac{\delta_0}{2} + 30^{\circ} + \Delta_i + \varepsilon_i\right) = n\sin(30^{\circ} + \varepsilon_i)$. Making the small angle approximation: $\sin\left(\frac{\delta_0}{2} + 30^{\circ}\right) + \cos\left(\frac{\delta_0}{2} + 30^{\circ}\right)(\Delta_i + \varepsilon_i) = n\sin 30^{\circ} + n\cos 30^{\circ}\cdot\varepsilon_i$

From the equation that holds for $60^{\circ}$ prism we get $\cos\left(\frac{\delta_0}{2} + 30^{\circ}\right)(\Delta_i + \varepsilon_i) = n\cos 30^{\circ}\cdot\varepsilon_i$

Averaging for all three angles we get $\langle\Delta_i\rangle = 0$, and therefore $n = 2\sin\left(\frac{\langle\delta_{\min}\rangle}{2} + 30^{\circ}\right)$

C.2. We will use the full length of the table to magnify the distances as much as possible. We will build the setup, as described in the sketch, so that in the absence of the prism, the laser beam will hit the screen (the chamber’s wall) perpendicularly. We will attach the prism holder base to the table using the adhesive tape. On it we will place the prism holder and the prism itself. We will rotate the prism to find the minimal deflection angle $\delta_{min}$. We will then repeat the measurement of $\delta_{min}$ for each corner of the prism.

The measured values are given in the table:

Calculation of the error in $\delta_{\min}$ :

Therefore, $\Delta\delta_{\min} = \cos^2\delta_{\min}\sqrt{\left(\frac{\Delta h}{L}\right)^2 + \left(\frac{h\Delta L}{L^2}\right)^2}$

Substituting the measured values we get

The error in the average value of the two angles is

From the table we get that the average value of $\delta_{\min}$ is $\langle\delta_{\min}\rangle = 50.50^{\circ}$

Therefore the refraction index of the prism is

And the error in $n$ : $\Delta n = 2\cos 55.25^{\circ}\cdot 0.5\Delta\langle\delta_{\min}\rangle = \cos 55.25^{\circ}\cdot 1\cdot 10^{-3} = 6\cdot 10^{-4}$

Thus: $n = 1.6433 \pm 0.0006$

As the laser wavelength may vary between lasers up to a standard deviation of $\pm 10\mathrm{nm}$, the value found in the literature is $n(\lambda \pm \Delta\lambda) = 1.6425 \pm 0.0007$.