$A$ step index fibre has a relative refractive index difference of $0.88\%$. What is the critical angle at the core-cladding interface in degrees?

Explain internal reflection and total internal reflection.

When a ray is refracted from one medium to another,the wavelength changes from $6000 \, \mathring A$ to $4000 \, \mathring A$. The critical angle for the interface will be

$A$ 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

$A$ ray of light travelling inside a rectangular glass block of refractive index $\sqrt{2}$ is incident on the glass-air surface at an angle of incidence of $45^{\circ}$. The refractive index of air is $1$. Under these conditions,the ray:

Consider the ray diagram for the refraction given below. The maximum value of angle $\theta$ for which the light suffers total internal reflection at the vertical surface is: