$A$ point source of light is placed at a distance $h$ below the surface of a large deep lake. What is the percentage of light energy that escapes directly from the water surface if the refractive index $\mu$ of the water is $4/3$ (in $\%$)? (Neglect partial reflection)

The minimum area of the disc needed to cut off all the light coming out of water $\left(\mu_{w} = \frac{4}{3}\right)$ from a point source of light placed $7 \ m$ below the water surface is: (in $\pi \ m^2$)

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7} \; m$. The refractive index of water is $\frac{4}{3}$. The area of the surface of water through which light from the bulb can emerge out is $x \,\pi \; m^{2}$. The value of $x$ is ...............

Assertion: The maximum refractive index of the liquid for total internal reflection of the ray passing through the prism as shown in the figure must be $\sqrt 2$.
Reason: Here,the critical angle is $45^o$.

$A$ ray of light is incident at an angle $i$ from a denser to a rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection is $r$ and the angle of refraction is $r'$. Then the critical angle $C$ will be:

At the interface between two materials having refractive indices $n_1$ and $n_2$,the critical angle for reflection of an electromagnetic wave is $\theta_{1C}$. The $n_2$ material is replaced by another material having refractive index $n_3$,such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2C}$. If $n_3 > n_2 > n_1$,$\frac{n_2}{n_3} = \frac{2}{5}$,and $\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$,then $\theta_{1C}$ is: