$A$ small coin is resting on the bottom of a beaker filled with liquid. $A$ ray of light from the coin travels up to the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?

At the interface between two materials having refractive indices $n_1$ and $n_2$,the critical angle for reflection of an electromagnetic wave is $\theta_{1C}$. The $n_2$ material is replaced by another material having refractive index $n_3$,such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2C}$. If $n_3 > n_2 > n_1$,$\frac{n_2}{n_3} = \frac{2}{5}$,and $\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$,then $\theta_{1C}$ is:

The graph between the sine of the angle of refraction $(\sin r)$ in medium $2$ and the sine of the angle of incidence $(\sin i)$ in medium $1$ is shown below. Based on this,which of the following statements is correct? (Given: $\tan 36^o \approx \frac{3}{4}$)

Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} \text{ m/s}$ and $2.0 \times 10^{8} \text{ m/s}$ respectively. The critical angle between them is.

The critical angle for diamond (refractive index $= 2$) is........$^o$

$A$ light ray falls on a glass surface of refractive index $\sqrt{3}$,at an angle of $60^{\circ}$. The angle between the refracted and reflected rays would be ....... $^o$.