The refractive indices of the material of the prism and the liquid are $1.56$ and $1.32$ respectively. Find the value of $\theta$ for total internal reflection at the interface.

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

$A$ diver at a depth of $12 \ m$ in water $\left(\mu = \frac{4}{3}\right)$ sees the sky in a cone of semi-vertical angle:

For the given incident ray as shown in the figure,what is the minimum refractive index of the prism required for total internal reflection of this ray?

Light guidance in an optical fiber can be understood by considering a structure comprising of a thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$. All rays with the angle of incidence $i$ less than a particular value $i_m$ are confined in the medium of refractive index $n_1$. The numerical aperture $(NA)$ of the structure is defined as $\sin i_m$.
$1.$ For two structures namely $S_1$ with $n_1=\sqrt{45}/4$ and $n_2=3/2$,and $S_2$ with $n_1=8/5$ and $n_2=7/5$,and taking the refractive index of water to be $4/3$ and that of air to be $1$,the correct option$(s)$ is(are):
$(A)$ $NA$ of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
$(B)$ $NA$ of $S_1$ immersed in a liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water
$(C)$ $NA$ of $S_1$ placed in air is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$
$(D)$ $NA$ of $S_1$ placed in air is the same as that of $S_2$ placed in water
$2.$ If two structures of the same cross-sectional area,but different numerical apertures $NA_1$ and $NA_2$ $(NA_2 < NA_1)$ are joined longitudinally,the numerical aperture of the combined structure is:
$(A)$ $\frac{NA_1 NA_2}{NA_1+NA_2}$ $(B)$ $NA_1+NA_2$ $(C)$ $NA_1$ $(D)$ $NA_2$

For a light ray to undergo total internal reflection, light must travel from $(i = \text{angle of incidence}, i_C = \text{critical angle})$