The minimum area of the disc needed to cut off all the light coming out of water $\left(\mu_{w} = \frac{4}{3}\right)$ from a point source of light placed $7 \ m$ below the water surface is: (in $\pi \ m^2$)

Assertion $(A)$: Optical fibres are widely used in communication networks. Reason $(R)$: Optical fibres are small in size, light in weight, flexible, and there is no scope for interference in them.

$A$ 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

$A$ ray of light in a liquid of refractive index $1.4$ approaches the boundary surface between the liquid and air at an angle of incidence whose sine is $0.8$. Which of the following statements is correct about the behavior of the light?

$A$ small source of light is $4 \ m$ below the surface of a liquid of refractive index $5/3$. In order to cut off all the light coming out of the liquid surface,the minimum diameter of the disc placed on the surface of the liquid is .......... $m$.

$A$ light ray travels from a medium with refractive index $n_1$ to another medium of refractive index $n_2$. If $n_1=2$ and $n_2=\sqrt{3}$, then find the critical angle. (in $^{\circ}$)