The wavelength of light, while it is passing through water, is $540 \ nm$. The refractive index of water is $\frac{4}{3}$. The wavelength of the same light when it is passing through a transparent medium having a refractive index of $\frac{3}{2}$ is . . . . . . $nm$.

$A$ ray of light is incident at the glass-water interface at an angle $i$. It emerges finally parallel to the surface of water. If the refractive index of water is $\mu_w = 4/3$,then the value of the refractive index of glass $\mu_g$ is:

The ratio of the thickness of plates of two transparent mediums $A$ and $B$ is $6 : 4$. If light takes equal time in passing through them,then the refractive index of $B$ with respect to $A$ will be:

Blue light travelling in a vacuum has a wavelength of $450 \ nm$. It enters a medium whose refractive index is $1.5$. What is its frequency in the medium? $(Speed of light in vacuum = 3 \times 10^8 \ m \ s^{-1})$

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:

$A$,$B$ and $C$ are parallel-sided transparent media of refractive indices $n_{1}$,$n_{2}$ and $n_{3}$ respectively. They are arranged as shown in the figure. $A$ ray is incident at an angle $i$ on the surface of separation of $A$ and $B$. After refraction into medium $B$,the ray grazes the surface of separation of media $B$ and $C$. Then,$\sin i$ equals to