$A$ ray of light travelling inside a rectangular glass block of refractive index $\sqrt{2}$ is incident on the glass-air surface at an angle of incidence of $45^{\circ}$. The refractive index of air is $1$. Under these conditions,the ray:

$A$ glass prism of refractive index $1.5$ is immersed in water (refractive index $4/3$) as shown in the figure. $A$ light beam incident normally on face $AB$ undergoes total internal reflection at face $BC$.

Four rays $1, 2, 3$ and $4$ are incident normally on the face $PQ$ of an isosceles prism $PQR$ with apex angle $\angle Q = 120^{\circ}$. The refractive indices of the material of the prism for the above rays $1, 2, 3$ and $4$ are $1.85, 1.95, 2.05$ and $2.15$ respectively and the surrounding medium is air. Then,the rays emerging from the face $QR$ are

In the figure shown,for an angle of incidence $45^{\circ}$ at the top surface,what is the minimum refractive index $\mu$ needed for total internal reflection at the vertical face?

The phenomenon of total internal reflection is observed when the angle of incidence is

At the interface between two materials having refractive indices $n_1$ and $n_2$,the critical angle for reflection of an electromagnetic wave is $\theta_{1C}$. The $n_2$ material is replaced by another material having refractive index $n_3$,such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2C}$. If $n_3 > n_2 > n_1$,$\frac{n_2}{n_3} = \frac{2}{5}$,and $\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$,then $\theta_{1C}$ is: