$A$ ray falls on a prism $ABC$ $(AB = BC)$ and travels as shown in the figure. The minimum refractive index of the prism material should be

The refracting angle of a prism is $A$ and the refractive index of the material of the prism is $\cot (A / 2)$. The angle of minimum deviation of the prism is

$A$ prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A,$ then the value of $A$ in terms of $\mu$ is

For an isosceles prism of angle $A$ and refractive index $\mu$,it is found that the angle of minimum deviation $\delta_m=A$. Which of the following options is/are correct?
[$A$] At minimum deviation,the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1=\left(i_1 / 2\right)$.
[$B$] For this prism the refractive index $\mu$ and the angle of prism $A$ are related as $A=\frac{1}{2} \cos ^{-1}(\mu / 2)$.
[$C$] For this prism,the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1=\sin ^{-1}\left[\sin A \sqrt{4 \cos ^2 \frac{A}{2}-1}-\cos A\right]$.
[$D$] For the angle of incidence $i_1=A$,the ray inside the prism is parallel to the base of the prism.

$A$ ray of monochromatic light is passing through an equilateral prism $(ABC)$ as shown in the figure. The refracted ray $(QR)$ is parallel to its base $(BC)$ and the angle of incidence $(i)$ is $50^\circ$. Then the angle of deviation $(\delta)$ is: (in $^\circ$)

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.