$A$ point source of light is placed $4 \, cm$ below the surface of water with a refractive index of $\mu = 5/3$. What is the minimum diameter of a disc that should be placed over the source to stop all light from coming out of the water? (in $m$)

$A$ ray of light is incident on one face of a $90^{\circ}$ prism and undergoes total internal reflection at the glass-air interface. If the angle of incidence is $45^{\circ}$,then the refractive index $n$ is:

$A$ right-angled isosceles prism is held on the surface of a liquid composed of miscible solvents $A$ and $B$ of refractive index $n_{A}=1.5$ and $n_{B}=1.3$,respectively. The refractive index of the prism is $n_{p}=1.5$ and that of the liquid is given by $n_{L}=C_{A} n_{A}+(1-C_{A}) n_{B}$,where $C_{A}$ is the percentage of solvent $A$ in the liquid. If $\theta_{C}$ is the critical angle at the prism-liquid interface,the plot which best represents the variation of the critical angle with the percentage of solvent is

The speed of light in media $M_{1}$ and $M_{2}$ are $1.5 \times 10^{8} \text{ m/s}$ and $2 \times 10^{8} \text{ m/s}$ respectively. $A$ ray travels from medium $M_{1}$ to the medium $M_{2}$ with an angle of incidence $\theta$. The ray suffers total internal reflection. Then the value of the angle of incidence $\theta$ is

Total internal reflection is possible when light rays travel:

In total internal reflection,when the angle of incidence is equal to the critical angle for the pair of media in contact,what will be the angle of refraction in degrees?