The prism angle of a prism is $40^{\circ}$. $A$ light ray is incident at $38^{\circ}$ and undergoes minimum deviation. The angle of minimum deviation is ........$^{\circ}$.

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

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. At what angle must a ray be incident on it so that it suffers minimum deviation (in $^{\circ}$)?

Explain the minimum deviation angle for a prism using the graph of deviation angle versus incidence angle.

$A$ parallel beam of light is incident on a thin prism with a prism angle of $\frac{4}{\pi}$ degrees. The refractive index of the prism is $1.5$. The focal length of the lens is $60 \ cm$. The coordinates of the converging point of the beam are:

$A$ light ray is incident on a prism of prism angle $60^{\circ}$ such that it undergoes minimum deviation. If the refractive index of the prism is $\sqrt{2}$,then the angle of incidence is .......$^{\circ}$.