$A$ 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

$A$ point source of light $S$ is placed at the bottom of a vessel containing a liquid of refractive index $\mu = 5/3$. $A$ person is viewing the source from above the surface. There is an opaque disc $D$ of radius $r = 1\, cm$ floating on the surface of the liquid. The centre of the disc lies vertically above the source $S$. The liquid from the vessel is gradually drained out through a tap. The maximum height $h$ of the liquid for which the source cannot be seen at all from above is: (in $, cm$)

What is the maximum acceptance angle at the entrance surface of an optical fiber,if $n_1$ and $n_2$ are the refractive indices of the core and cladding,respectively?

The phenomenon of total internal reflection is observed when the angle of incidence is

Let the refractive index of a denser medium with respect to a rarer medium be $n_{12}$ and its critical angle be $\theta_C$. At an angle of incidence $A$ when light is travelling from a denser medium to a rarer medium,a part of the light is reflected and the rest is refracted,and the angle between the reflected and refracted rays is $90^o$. Angle $A$ is given by

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: