Does Not Look Like What?

From the post "A Bloom Crosses Over" dated 10 Aug 2015,

$\cfrac{ tan(\theta_1)}{tan(\theta_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

$\varepsilon_2\gt\varepsilon_1$

then,

$\theta_2\lt\theta_1$

$\Delta\theta=\theta_2-\theta_1\lt0$

In all the previous diagrams $\Delta\theta\gt0$.

If $\Delta\theta^{\,'}=-\Delta\theta$

where $\varepsilon_2\gt\varepsilon_1$.