$A$ light wave travelling linearly in a medium of dielectric constant $4$ is incident on the horizontal interface separating the medium from air. The angle of incidence for which the total intensity of the incident wave will be reflected back into the same medium is (Given: relative permeability of the medium $\mu_{r} = 1$) (in $^{\circ}$)

The wavelengths of given light waves in air and in a medium are $6000 \ Å$ and $4000 \ Å$ respectively. The critical angle is

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:

$A$ glass prism $(\mu = 1.5)$ is dipped in water $(_a\mu_w = 4/3)$ as shown in the figure. $A$ light ray is incident normally on the surface $AB$. It reaches the surface $BC$ after being totally internally reflected at the surface $AC$,if:

In total internal reflection,when the angle of incidence is equal to the critical angle for the pair of media in contact,what will be the angle of refraction in degrees?

Light travels in two media $A$ and $B$ with speeds $1.8 \times 10^8 \ m/s$ and $2.4 \times 10^8 \ m/s$ respectively. Then the critical angle between them is