Angle of deviation $(\delta)$ by a prism (refractive index = $\mu$ and supposing the angle of prism $A$ to be small) can be given by

If a prism having refractive index $\sqrt{3}$ has an angle of minimum deviation equal to the refracting angle of the prism,then the refracting angle of the prism is......$^o$

The refractive index of the material of a glass prism is $\sqrt{3}$. If the angle of minimum deviation is equal to the angle of the prism,then the angle of the prism is:
$\left(\cos 30^{\circ} = \frac{\sqrt{3}}{2} = \sin 60^{\circ}, \sin 30^{\circ} = \frac{1}{2} = \cos 60^{\circ}\right)$ (in $^{\circ}$)

$A$ ray of light is incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$. It produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is

$A$ ray $OP$ of monochromatic light is incident on the face $AB$ of prism $ABCD$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$,which of the following is (are) correct?
$(A)$ The ray gets totally internally reflected at face $CD$
$(B)$ The ray comes out through face $AD$
$(C)$ The angle between the incident ray and the emergent ray is $90^{\circ}$
$(D)$ The angle between the incident ray and the emergent ray is $120^{\circ}$

For the angle of minimum deviation of a prism to be equal to its refracting angle,the prism must be made of a material whose refractive index is: