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 \(\varepsilon_2\gt\varepsilon_1\).