On the hypotenuse of a right-angled prism $(30^{\circ}-60^{\circ}-90^{\circ})$ with a refractive index of $1.50$,a drop of liquid is placed as shown in the figure. Light is allowed to fall normally on the short face of the prism. In order for the ray of light to undergo total internal reflection at the hypotenuse,what is the maximum value of the refractive index of the liquid?

When a ray of light emerges from a block of glass,the critical angle is

Assertion $(A)$: Optical fibres are widely used in communication networks. Reason $(R)$: Optical fibres are small in size, light in weight, flexible, and there is no scope for interference in them.

$A$ 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

Assertion: Optical fibres are used for telecommunication.
Reason: Optical fibres are based on the phenomenon of total internal reflection.

$A$ light ray traveling in a glass medium is incident on a glass-air interface at an angle of incidence $\theta$. The reflected $(R)$ and transmitted $(T)$ intensities,both as functions of $\theta$,are plotted. The correct sketch is: