$A$ ray of light travelling through glass of refractive index $\sqrt{2}$ is incident on a glass-air boundary at an angle of incidence of $45^{\circ}$. If the refractive index of air is $1$,then the angle of refraction will be: $[\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 90^{\circ} = 1]$ (in $^{\circ}$)

Is the statement "Angle made by the incident ray with the reflecting surface is the angle of incidence" correct?

An observer can see through a small hole on the side of a jar (radius $15\, cm$) at a point at a height of $15\, cm$ from the bottom (see figure). The hole is at a height of $45\, cm$. When the jar is filled with a liquid up to a height of $30\, cm$,the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is $N/100$,where $N$ is an integer,the value of $N$ is $.....$

Time taken by light to travel in two different materials $A$ and $B$ of refractive indices $\mu_{A}$ and $\mu_{B}$ of the same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2}-t_{1}=5 \times 10^{-10} \text{ s}$ and the ratio of $\mu_{A}$ to $\mu_{B}$ is $1:2$. Then the thickness of the material,in meters,is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

Monochromatic light is refracted from air into glass of refractive index $\mu$. The ratio of the wavelength of incident and refracted waves is

$A$ beaker containing liquid is placed on a table,underneath a microscope which can be moved along a vertical scale. The microscope is focused,through the liquid,onto a mark on the table when the reading on the scale is $a$. It is next focused on the upper surface of the liquid and the reading is $b$. More liquid is added and the observations are repeated; the corresponding readings are $c$ and $d$. The refractive index of the liquid is