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

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

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \, cm$. What is the area of the surface of water through which light from the bulb can emerge out? The refractive index of water is $1.33$. (Consider the bulb to be a point source.)

$A$ light ray falls on a rectangular glass slab as shown in the figure. If total internal reflection occurs at the vertical face of the slab at point $B$,the refractive index of glass is

$A$ girl looks through a circular glass slab (refractive index $1.5$) of thickness $20 \,mm$ and diameter $60 \,cm$ to the bottom of a swimming pool. The refractive index of water is $1.33$ (or $4/3$). The bottom surface of the slab is in contact with the water surface. The depth of the swimming pool is $6 \,m$. The area of the bottom of the swimming pool that can be seen through the slab is approximately ..............$m^2$.

$A$ 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: