$A$ fish inside water sees the outside world through a circular horizon. If the refractive index of water is $4/3$ and the fish is $12 \, cm$ below the water surface,then the radius of this circle is $= ..... cm$.

The critical angle of a medium with respect to air is $45^o$. The refractive index of the medium 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

$A$ 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

When a ray is refracted from one medium to another,the wavelength changes from $6000 \, \mathring A$ to $4000 \, \mathring A$. The critical angle for the interface will be

For a light ray to undergo total internal reflection ($i =$ angle of incidence,$i_c =$ critical angle):