$A$ ray of light is incident at $30^{\circ}$ from a medium of refractive index $2$ into a medium of refractive index $1$. The angle of refraction is: (in $^{\circ}$)

The velocity of light in glass having a refractive index of $1.5$ is $2 \times 10^8 \ m/s$. What is the refractive index of a liquid in which the velocity of light is $2.50 \times 10^8 \ m/s$?

$A$ glass slab consists of thin uniform layers of progressively decreasing refractive indices $(RI)$ such that the $RI$ of any layer is $\mu - m \Delta \mu$. Here,$\mu$ and $\Delta \mu$ denote the $RI$ of the $0^{\text{th}}$ layer and the difference in $RI$ between any two consecutive layers,respectively. The integer $m = 0, 1, 2, 3, \ldots$ denotes the number of the successive layers. $A$ ray of light from the $0^{\text{th}}$ layer enters the $1^{\text{st}}$ layer at an angle of incidence of $30^{\circ}$. After undergoing the $m^{\text{th}}$ refraction,the ray emerges parallel to the interface. If $\mu = 1.5$ and $\Delta \mu = 0.015$,the value of $m$ is:

$A$ light ray incident along a vector $\vec{AO}$ $(\vec{AO} = 2\hat{i} - 3\hat{j})$ emerges along a vector $\vec{OB}$ $(\vec{OB} = C\hat{i} - 4\hat{j})$ as shown in the figure. The value of $C$ is . . . . . . .

When light travels from glass to air,the incident angle is ${\theta _1}$ and the refracted angle is ${\theta _2}$. The true relation is:

If $_i{\mu _j}$ represents the refractive index when a light ray goes from medium $i$ to medium $j,$ then the product $_2{\mu _1} \times {\,_3}{\mu _2} \times {\,_4}{\mu _3}$ is equal to