The critical angle for total internal reflection of light going from medium $I$ to medium $II$ is given by the relation $\tan i_C = \frac{5}{9}$. The refractive index of medium $I$ with respect to medium $II$ is:

$A$ point source of light is placed at a depth $h = 0.5 \, m$ below the surface of a liquid $(\mu = \frac{5}{4})$. Then,the fraction of light energy that escapes directly from the liquid surface is:

The speed of light in media $M_{1}$ and $M_{2}$ are $1.5 \times 10^{8} \text{ m/s}$ and $2 \times 10^{8} \text{ m/s}$ respectively. $A$ ray travels from medium $M_{1}$ to the medium $M_{2}$ with an angle of incidence $\theta$. The ray suffers total internal reflection. Then the value of the angle of incidence $\theta$ is

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

$A$ light ray is incident from a denser medium to a rarer medium at an angle of incidence $i$. If the reflected and refracted rays are mutually perpendicular,then the critical angle for the interface is:

$A$ ray of light travels in the fashion as shown in the figure. After passing through water,the ray grazes along the water-air interface. The value of $\mu_g$ in terms of $i$ is