$A$ ray of light is incident at an angle '$i$' on one face of a thin 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:

$A$ light ray is incident on a prism at an angle of $60^{\circ}$ and the refracting angle of the prism is $30^{\circ}$. The emergent ray makes an angle of $30^{\circ}$ with the incident ray. The refractive index of the prism 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 $^{\circ}$)

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 is:

$A$ graph is plotted between angle of deviation $(\delta)$ and angle of incidence $(i)$ for a prism. The nearly correct graph is

$A$ ray of light is incident normally on the diagonal face of a right-angled prism as shown in the figure. If $\theta = 37^o$ and the refractive index of the prism is $\mu = 5/3$,then the angle of deviation is......$^o$ $(\sin37^o = 3/5)$