For a prism,the angle of the prism is $60^{\circ}$ and the refractive index is $\sqrt{7/3}$. The minimum possible angle of incidence so that the light ray is refracted from the second surface is (in $^{\circ}$)

Consider the following statements for the refraction of light through a prism,when the angle of deviation is minimum.
$(A)$ The refracted ray inside the prism becomes parallel to the base.
$(B)$ Larger angle prisms provide smaller angle of minimum deviation.
$(C)$ The angle of incidence and the angle of emergence become equal.
$(D)$ There are always two sets of angles of incidence for which the deviation will be the same,except at the minimum deviation setting.
$(E)$ The angle of refraction becomes double the prism angle.
Choose the correct answer from the options given below.

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$. The emergent ray makes an angle of $30^{\circ}$ with the incident ray. The refractive index of the prism is

Calculate the angle of minimum deviation for an equilateral triangular prism of refractive index $\sqrt{3}$. (in $^{\circ}$)

$A$ monochromatic beam of light is incident at $60^{\circ}$ on one face of an equilateral prism of refractive index $n$ and emerges from the opposite face making an angle $\theta(n)$ with the normal (see the figure). For $n=\sqrt{3}$ the value of $\theta$ is $60^{\circ}$ and $\frac{d \theta}{d n}=m$. The value of $m$ is

$A$ prism of refractive index $\sqrt{2}$ and refracting angle $A$ produces a minimum deviation $D_m$. $A$ ray incident on one face at an angle of incidence $i = 45^{\circ}$ undergoes minimum deviation. The values of $A$ and $D_m$ are respectively: