# The figure shows total internal reflection taking place in three materials of refractive indices $n_1,n_2$ and $n_3$ where $\theta _{\large c_1},\theta _{\large c_2},\theta _{\large c_3}$ are critical angles in the respective mediums. Then

$(a)\;n_1 > n_2 > n_3 \\ (b)\;n_3 < n_1 < n_2 \\ (c)\;n_1 < n_2 < n_3 \\ (d)\;n_2 < n_1 < n_3$

$\sin \theta _c =\large\frac{1}{\mu}$
Since third material has maximum critical angle, 2nd material least
$\theta _{\Large c_3} > \theta _{\Large c_1} > \theta _{\Large c_2}$
We have $n_3 < n_1 < n_2$
Hence b is the correct answer.