$A$ diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\frac{4}{3}$,and the diver's eyes are $15 \, cm$ below the surface of water. Then the radius of the circle is

An optical fibre consists of a core of refractive index $\mu_1$ surrounded by a cladding of refractive index $\mu_2 < \mu_1$. $A$ beam of light enters from air at an angle $\alpha$ with the axis of the fibre. The highest angle $\alpha$ for which the ray can travel through the fibre is:

$A$ 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 $\mu s$)?

White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected,then the emerging ray in air contains:

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:
$(A)$ $\frac{NA_1 NA_2}{NA_1+NA_2}$ $(B)$ $NA_1+NA_2$ $(C)$ $NA_1$ $(D)$ $NA_2$

If $\mu_1$ and $\mu_2$ are the refractive indices of the materials of core and cladding of an optical fibre,then the loss of light due to its leakage can be minimised by having