$A$ light ray is incident from a denser medium to a rarer medium at an angle of incidence $i$. If the reflected and refracted rays are mutually perpendicular,then the critical angle for the interface is:

$A$ light ray is incident normally on one face of a right-angled prism and travels parallel to the base inside the prism. If the refractive index of the prism is $\mu$,what should be the angle $\alpha$ that the hypotenuse makes with the base so that the ray undergoes total internal reflection at the hypotenuse?

Initially,$AB$ is the interface of a slab of refractive index $\mu$ and air. $A$ ray of light strikes $AB$ at the critical angle $\theta_c$. Now,a slab $ABCD$ of refractive index $\mu_1$ is placed on top such that total internal reflection takes place at the surface $CD$ (and not at the surface $AB$). What should be the value of $\mu_1$?

In an optical fibre, is the refractive index of the core greater than that of the cladding, or smaller?

$A$ transparent solid cylindrical rod has a refractive index of $\frac{4}{\sqrt{3}}$. It is surrounded by a medium of refractive index $2$. $A$ 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

$A$ 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: