$A$ spider and a fly are on opposite sides of the surface of a glass sphere. What is the maximum solid angle $\Omega$ within which the fly can be seen by the spider?
Given: The dimensions of the spider and the fly are very small with respect to the sphere. The refractive index of the glass is $\mu_g = \sqrt{2}$.

Total internal reflection of light is possible when light enters from

$A$ ray of light from a denser medium strikes a rarer medium. The angle of reflection is $r$ and that of refraction is $r'$. The reflected and refracted rays make an angle of $90^{\circ}$ with each other. The critical angle will be

$A$ 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 $n_2 = 1$

The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium into air,the angle of incidence,at which it will be totally internally reflected,is ......... $^o$

$A$ 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 ..............$m^2$.