The critical angle for total internal reflection of light going from medium $I$ to medium $II$ is given by the relation $\tan i_C = \frac{5}{9}$. The refractive index of medium $I$ with respect to medium $II$ is:

$A$ 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

$A$ transparent cube of side $d$,made of a material of refractive index $\mu_2$,is immersed in a liquid of refractive index $\mu_1$ (where $\mu_1 < \mu_2$). $A$ ray is incident on the face $AB$ at an angle $\theta$ (as shown in the figure). Total internal reflection takes place at point $E$ on the face $BC$. The angle $\theta$ must satisfy:

Light travels a distance '$x$' in time '$t_0$' in air and '$4x$' in time '$t_1$' in another denser medium. The critical angle for this medium is

Optical fibres are related with

$A$ 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