Angle measurement of Polaris

The Snell's law for the refraction on the glass gives $\begin{aligned} n_{\text{outside}} \sin \alpha &= n_\text{glass} \sin \gamma = n_{\text{inside}} \sin \beta \\ \Rightarrow \quad \beta &= \arcsin \left[ \frac{n_{\text{outside}}}{n_{\text{inside}}} \sin \alpha \right] \end{aligned}$ with the refractive indices $n_\text{glass}$ of the glass pane as well as $n_ {\text{inside}}$ and $n_ {\text{outside}}$ of the air inside and outside. Although the air inside and outside has the same composition and pressure $p \approx 1 \, \text{bar} = 10^5 \, \text{Pa}$ , there is a difference in the temperature difference the density $\rho$ . With the ideal gas law follows $\begin{aligned} & & p &= R^\ast \rho T \qquad \Leftrightarrow \qquad \rho = \frac{p}{R^\ast T} \\ \Rightarrow & & n &= \sqrt{\varepsilon} = \sqrt{1 + \chi} = \sqrt{1 + c \rho} = \sqrt{1 + \frac{c p}{R^\ast T}} \end{aligned}$ Here we have use the fact that the air molecules show almost no magnetism at room temperature ( $\mu \approx 1$ ), so that the refractive index of the air is only determined by dielectric properties ( $n = \sqrt{\varepsilon \mu} \approx \sqrt{\varepsilon}$ ). Therefore, $\begin{aligned} \beta &= \arcsin \left[ \frac{n_{\text{outside}}}{n_{\text{inside}}} \sin \alpha \right] \\ &= \arcsin \left[ \sqrt{\frac{1 + \frac{c p}{R^\ast T_\text{outside}}}{1 + \frac{c p}{R^\ast T_\text{inside}}}} \sin \alpha \right]\\ &\approx \arcsin \left[ \sqrt{1 + \frac{c p}{R^\ast T_\text{outside}} - \frac{c p}{R^\ast T_\text{inside}}} \sin \alpha \right]\\ &\approx \arcsin \left[ \left(1 + \frac{1}{2} \frac{c p}{R^\ast} \left[\frac{1}{T_\text{outside}} - \frac{1}{T_\text{inside}} \right] \right) \sin \alpha \right]\\ &\approx \alpha + \frac{1}{\sqrt{1 - (\sin \alpha)^2}} \cdot \frac{c p}{2 R^\ast} \left[\frac{1}{T_\text{outside}} - \frac{1}{T_\text{inside}} \right] \sin \alpha \end{aligned}$ Here we have taken advantage of the fact that the refractive index of air differs only slightly from 1, so we used the approximations $\begin{aligned} \frac{1}{1 + x} &\approx 1 - x \\ \sqrt{1 + x} &\approx 1 + \frac{1}{2} x \end{aligned}$ for $x \ll 1$ and also made an linear Taylor expansion for the arcus sine.

For the calculation of the angle $\beta$ we have to know the angle height $\alpha$ of the polar star. Venice lies at the latitude $\phi = 45^\circ$ , so that the position of the polar star also results to $\alpha \approx 45 ^\circ$ , since this star is located at the north pole of the sky. Therefore, $\sin \alpha = \frac{1}{\sqrt {2}}$ and thus $\begin{aligned} \delta = \beta- \alpha &\approx \frac{c p}{2 R^\ast} \left[\frac{1}{T_\text{outside}} - \frac{1}{T_\text{inside}} \right] \\ &= \frac{10^{-3} \cdot 10^5}{2 \cdot 300} \left[\frac{1}{270} - \frac{1}{293} \right] \\ &\approx 4.8 \cdot 10^{-5} \,\text{rad}\\ &\approx 0.0028 ^\circ \\ &\approx 10'' \end{aligned}$