$A$ light ray is incident on a prism of angle $A = 30^{\circ}$ at an angle of incidence of $60^{\circ}$. If the emergent ray makes an angle of $30^{\circ}$ with the incident ray,then find the refractive index of the prism $:-$

$A$ monochromatic light is incident from air on a refracting surface of a prism of angle $A = 75^{\circ}$ and refractive index $n_0 = \sqrt{3}$. The other refracting surface of the prism is coated by a thin film of material of refractive index $n$ as shown in the figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $\theta \leq 60^{\circ}$. The value of $n^2$ is:

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 $^{\circ}$)

$A$ light ray enters through a right-angled prism at point $P$ with an angle of incidence of $30^{\circ}$ as shown in the figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:

$A$ prism is made up of a material with a refractive index of $\sqrt{3}$. The angle of the prism is $A$. If the angle of minimum deviation is equal to the angle of the prism, then the value of $A$ is: (in $^{\circ}$)

$A$ ray of light incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$, produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is $(\sin 90^{\circ} = 1, \sin 30^{\circ} = 0.5, \sin 45^{\circ} = 1/\sqrt{2})$.