To solve this problem, we need to understand the condition given:$\\$The reflected and refracted rays are perpendicular to each other.$\\$This condition is satisfied at Brewster's angle, where the angle of incidence $i_B$ is such that:$\\ i_B + r = 90^\circ \\$Here, $r$ is the angle of refraction. Since $i_B + r = 90^\circ,$ we have:$\\ r = 90^\circ - i_B \\$Using Snell's law:$\\ n_1 \sin i_B = n_2 \sin r \\$For air, $n_1 = 1$ and for the material, $n_2 = \mu.$ Therefore,$\\ \sin i_B = \mu \sin(90^\circ - i_B) \\$Since $\sin(90^\circ - i_B) = \cos i_B,$ we have:$\\ \sin i_B = \mu \cos i_B \\$Dividing both sides by $\cos i_B,$ we get:$\\ \tan i_B = \mu \\$Thus, the angle of incidence $i_B$ is given by:$\\ i_B = \tan^{-1}(\mu) \\$This means that the reflected light is completely polarized when the angle of incidence is Brewster's angle.$\\$At Brewster's angle, the reflected light is polarized with its electric vector perpendicular to the plane of incidence.$\\$Therefore, the correct option is:$\\$Option 2: Reflected light is polarised with its electric vector perpendicular to the plane of incidence.