The refractive index of the material of a glass prism is $\sqrt{3}$. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism (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

In the given figure,the face $AC$ of the equilateral prism is immersed in a liquid of refractive index $n$. For an incident angle of $60^{\circ}$ at the side $AC$,the refracted light beam just grazes along the face $AC$. The refractive index of the liquid is $n = \frac{\sqrt{x}}{4}$. The value of $x$ is (Given refractive index of glass $= 1.5$)

Minimum deviation is observed with a prism having angle of prism $A$,angle of deviation $\delta$,angle of incidence $i$,and angle of emergence $e$. We then have generally

$A$ ray of light is incident at an angle '$i$' on one face of a thin angled prism. The ray emerges normally from the other face. If the refractive index of the glass prism is '$n$' and the angle of the prism is '$A$',then the value of '$i$' is:

Define the angle of minimum deviation.