$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \; cm$. What is the area (in $m^2$) of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

The speed of light in two media $c_1$ and $c_2$ are $1.5 \times 10^8 \ m/s$ and $2 \times 10^8 \ m/s$ respectively. If the light undergoes total internal reflection,the critical angle between the two media 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 ..............$m^2$.

$A$ light wave enters from medium $1$ to medium $2$. Its velocity in the $2^{nd}$ medium is double that in the $1^{st}$ medium. For total internal reflection,the angle of incidence must be greater than.......$^o$.

The figure shows a ray incident at an angle $i = \pi /3.$ If the plot drawn shows the variation of $|r - i|$ versus $\frac{\mu_1}{\mu_2} = k,$ ($r =$ angle of refraction) choose the wrong alternative.

The critical angle for light going from medium $A$ into medium $B$ is $\theta$. The speed of light in medium $A$ is $V_A$. What is the speed of light in medium $B$?