$A$ rectangular glass block is placed on a printed paper lying on a horizontal surface. Find the minimum refractive index of the glass for which the letters on the paper are not visible from any of the vertical sides of the block.

An optical fibre consists of a core of refractive index $\mu_1$ surrounded by a cladding of refractive index $\mu_2 < \mu_1$. $A$ beam of light enters from air at an angle $\alpha$ with the axis of the fibre. The highest angle $\alpha$ for which the ray can travel through the fibre is:

Three transparent media of refractive indices $\mu_1, \mu_2, \mu_3$,respectively,are stacked as shown below. $A$ ray of light follows the path shown. No light enters the third medium. Then,

The critical angle for a denser-rarer interface is $45^{\circ}$. The speed of light in the rarer medium is $3 \times 10^8 \, m/s$. The speed of light in the denser medium is:

Let the refractive index of a denser medium with respect to a rarer medium be $n_{12}$ and its critical angle be $\theta_C$. At an angle of incidence $A$ when light is travelling from a denser medium to a rarer medium,a part of the light is reflected and the rest is refracted,and the angle between the reflected and refracted rays is $90^o$. Angle $A$ is given by

Assertion: The refractive index of diamond is $\sqrt{6}$ and that of liquid is $\sqrt{3}$. If the light travels from diamond to the liquid,it will be totally internally reflected when the angle of incidence is $30^{\circ}$.
Reason: $\mu = \frac{1}{\sin C}$,where $\mu$ is the refractive index of diamond with respect to the liquid.