$A$ light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass,if total internal reflection is to occur at the vertical face,is equal to

Four rays $1, 2, 3$ and $4$ are incident normally on the face $PQ$ of an isosceles prism $PQR$ with apex angle $\angle Q = 120^{\circ}$. The refractive indices of the material of the prism for the above rays $1, 2, 3$ and $4$ are $1.85, 1.95, 2.05$ and $2.15$ respectively and the surrounding medium is air. Then,the rays emerging from the face $QR$ are

The figure shows a transparent block of side lengths $a$ and $a$. The third dimension of the block is negligible. $A$ point source $S$,which can emit light in all directions,can move inside the block. It is desired that no light from $S$ should pass through the segment $AB$. The region in which $S$ must be present to satisfy this condition is shown by the shaded region. Choose the correct option.

What should be the maximum acceptance angle at the air-core interface of an optical fibre if $n_1$ and $n_2$ are the refractive indices of the core and the cladding,respectively?

$A$ ray of light is incident at an angle $i$ from a denser to a rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection is $r$ and the angle of refraction is $r'$. Then the critical angle $C$ will be:

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: