Suppose the ray travels from point $A$ in medium 1 to point $B$ in medium 2, crossing the interface at position $x$. The perpendicular distances of $A$ and $B$ to the interface are $a$ and $b$, and the projected separation along the interface is $d$. By the Pythagorean theorem the two legs have lengths $\sqrt{a^2+x^2}$ and $\sqrt{b^2+(d-x)^2}$; dividing by the speeds $v_1$, $v_2$ in each medium and adding gives the total time $T(x)$.