The refractive indices of glass and water with respect to air are $3/2$ and $4/3$ respectively. The refractive index of glass with respect to water is:

If $\hat{i}$ denotes a unit vector along the incident light ray,$\hat{r}$ a unit vector along the refracted ray into a medium of refractive index $\mu$,and $\hat{n}$ a unit vector normal to the boundary of the medium directed towards the incident medium,then the law of refraction is:

The velocity of light in diamond is $\left(\frac{5}{12}\right)$ times that in air. The velocity of light in water is $\left(\frac{3}{4}\right)$ times that in air. Find the angle of incidence for a ray of light travelling from water to diamond,given that the angle of refraction $r = 30^{\circ}$ and $\sin 30^{\circ} = \frac{1}{2}$.

$A$ girl standing at point $P$ on a beach wishes to reach a point $Q$ in the sea as quickly as possible. She can run at $6 \, kmh^{-1}$ on the beach and swim at $4 \, kmh^{-1}$ in the sea. She should take the path

$A$ 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:

For a colour of light,the wavelength in air is $6000 \ \mathring A$ and in water,the wavelength is $4500 \ \mathring A$. Then the speed of light in water will be: