The angle of minimum deviation is equal to the angle of the equilateral prism. At what angle of incidence $i$ will this minimum deviation occur?

$A$ glass prism has a right-triangular cross-section $ABC$,with $\angle A=90^{\circ}$. $A$ ray of light parallel to the hypotenuse $BC$ and incident on the side $AB$ emerges grazing the side $AC$. Another ray,again parallel to the hypotenuse $BC$,incident on the side $AC$ suffers total internal reflection at the side $AB$. Which one of the following must be true about the refractive index $\mu$ of the material of the prism?

If the refractive index of the material of a prism is $\cot(A/2)$,where $A$ is the angle of the prism,then the angle of minimum deviation will be:

$A$ 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:

$A$ thin glass prism has a refractive index of $1.5$. The correct relation between the angle of minimum deviation $(\delta_m)$ and the angle of the prism $(A)$ is given by the formula for a thin prism. If the angle of refraction is $r$, find the correct relation between the angle of minimum deviation $(\delta_m)$ and the angle of refraction $(r)$ for a thin prism.

$A$ ray of light is incident on the first face of an equilateral glass prism at an angle of incidence '$i$'. The emergent ray just grazes along the adjacent face. The refractive index of the prism is $\sqrt{2}$. The value of the angle of incidence at the first face is $(\sin 45^{\circ} = 1/\sqrt{2}, \sin 90^{\circ} = 1)$.