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.

The refractive indices of the material of the prism and the liquid are $1.56$ and $1.32$ respectively. Find the value of $\theta$ for total internal reflection at the interface.

In the figure,$ABC$ is the cross-section of a right-angled prism and $BCDE$ is the cross-section of a glass slab. The value of $\theta$ so that light incident normally on the face $AB$ does not cross the face $BC$ is (given $\sin^{-1}(3/5) = 37^o$):

$A$ transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$. It is surrounded by air. $A$ light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected,then the emerging ray in air contains:

$A$ planar structure of length $L$ and width $W$ is made of two different optical media of refractive indices $n_1=1.5$ and $n_2=1.44$ as shown in the figure. If $L \gg W$,a ray entering from end $AB$ will emerge from end $CD$ only if the total internal reflection condition is met inside the structure. For $L = 9.6 \ m$,if the incident angle $\theta$ is varied,the maximum time taken by a ray to exit the plane $CD$ is $t \times 10^{-9} \ s$,where $t$ is. . . . . . . [Speed of light $c = 3 \times 10^8 \ m/s$]