$A$ ray of light originates from inside a glass slab and is incident on its inner surface at an angle $\theta$ as shown below. In this experiment,the location $x$ of the spot where the ray hits the screen is recorded. Which of the following correctly shows the plot of variation of $x$ with the angle $\theta$?

$A$ light ray travels from a medium with refractive index $n_1$ to another medium of refractive index $n_2$. If $n_1=2$ and $n_2=\sqrt{3}$, then find the critical angle. (in $^{\circ}$)

$A$ point source $S$ is placed at the bottom of different layers as shown in the figure. The refractive index of the bottom-most layer is $\mu_0$. The refractive index of any other upper layer is $\mu(n) = \mu_0 - \frac{\mu_0}{4n - 18}$,where $n = 1, 2, \dots$. $A$ ray of light with an angle of incidence $i$ slightly more than $30^\circ$ starts from the source $S$. Total internal reflection takes place at the upper surface of a layer having $n$ equal to

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:

Assertion $(A)$: Propagation of light through an optical fibre is due to total internal reflection taking place at the core-clad interface.
Reason $(R)$: Refractive index of the material of the core of the optical fibre is greater than that of air.

$A$ spider and a fly are on opposite sides of the surface of a glass sphere. What is the maximum solid angle $\Omega$ within which the fly can be seen by the spider?
Given: The dimensions of the spider and the fly are very small with respect to the sphere. The refractive index of the glass is $\mu_g = \sqrt{2}$.