 Solution

The distance from the origin to the incident straight light ray is $d_a$ (taking $d_a\ge 0$). At its intersection with the sphere, let $i$ and $t$ be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives $d_a=r_b\sin i$. By Snell's law,The refracted straight light ray inside the sphere has distance $d_b=r_b\sin t$ from the origin. ThereforeThe angle ${\theta }_b$ describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires $|d_a|<r_b$. A signed convention for $d_a$ gives the same line equation after choosing the orientation of its normal vector consistently.