When a rectangular metal tank is filled to the top with an unknown liquid,an observer with eyes level with the top of the tank can just see the corner $E$. $A$ light ray that refracts towards the observer at the top surface of the liquid is shown. The refractive index of the liquid 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

$A$ point source of light is placed $4 \; m$ below the surface of water of refractive index $5/3$. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is ... $m$.

$A$ normally incident ray is reflected at an angle of $90^o$. The value of the critical angle is.......$^o$

Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} \text{ m/s}$ and $2.0 \times 10^{8} \text{ m/s}$ respectively. The critical angle between them is.

The critical angle is $\theta$ for light going from medium $P$ to medium $Q$. If the speed of light in medium $P$ is $V_{P}$,then the speed of light in medium $Q$ is: