$A$ light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive index $\mu = 1.5$) making an incidence angle of $60^o$ on the lateral surface,so that it undergoes a total internal reflection. How much time would it take to traverse the straight fiber of length $1 \ km$ (in $\mu s$)?

$A$ 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:

The speed of light in media $A$ and $B$ are $2.0 \times 10^{10} \, cm/s$ and $1.5 \times 10^{10} \, cm/s$ respectively. $A$ ray of light enters from medium $B$ to $A$ at an incident angle $\theta$. If the ray suffers total internal reflection,then:

What is the least radius through which an optical fiber of core diameter $0.05 \, mm$ may be bent (as shown in figure) without serious loss of light (in $, mm$)? The refractive index of the core is $1.6$ and that of cladding is $1.5$.

Three transparent media of refractive indices $\mu_1, \mu_2, \mu_3$,respectively,are stacked as shown below. $A$ ray of light follows the path shown. No light enters the third medium. Then,

$A$ wide slab consisting of two media of refractive indices $n_1$ and $n_2$ is placed in air as shown in the figure. $A$ ray of light is incident from medium $n_1$ to $n_2$ at an angle $\theta$,where $\sin \theta$ is slightly larger than $1/n_1$. Take the refractive index of air as $1$. Which of the following statement$(s)$ is(are) correct?
$(A)$ The light ray enters air if $n_2 = n_1$
$(B)$ The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 < n_1$
$(C)$ The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 > n_1$
$(D)$ The light ray is reflected back into the medium of refractive index $n_1$ if $n_2 = 1$