The bottom of a container filled with liquid appears slightly raised because of

$A$ 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 passes through four transparent media with refractive indices $\mu_1, \mu_2, \mu_3$ and $\mu_4$ as shown in the figure. All surfaces are parallel to each other. If the emergent ray $CD$ is parallel to the incident ray $AB$,then:

$A$ ray of light falls on a transparent glass slab with a refractive index (relative to air) of $1.62$. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is

An observer can see through a pin-hole the top end of a thin rod of height $h$,placed as shown in the figure. The beaker height is $3h$ and its radius is $h$. When the beaker is filled with a liquid up to a height $2h$,the observer can see the lower end of the rod. Then the refractive index of the liquid is

If $\varepsilon_{0}$ and $\mu_{0}$ are the permittivity and permeability of free space and $\varepsilon$ and $\mu$ are the corresponding quantities for a medium,then the refractive index of the medium is: