The refracting angle of a glass prism is $30^{\circ}$. $A$ ray is incident on one of the faces and is perpendicular to it. The angle of deviation $\delta$ between the incident ray and the ray that leaves the prism is (Refractive index of glass $= 1.5$) (Given: $\sin(30^{\circ}) = 0.5, \sin(48.6^{\circ}) = 0.75$). (in $^{\circ}$)

The angle of deviation $\delta$ produced by a prism of refractive index $\mu$ and small refracting angle $A$ is given by:

Two equilateral-triangular prisms $P_1$ and $P_2$ are kept with their sides parallel to each other,in vacuum,as shown in the figure. $A$ light ray enters prism $P_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $P_2$. If the respective refractive indices of $P_1$ and $P_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$,then $\theta = \sin^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$,where the value of $\beta$ is:

$A$ yellow light undergoes minimum deviation through a prism. If $i_1$ and $i_2$ are the angle of incidence and the angle of emergence respectively,then:

$A$ prism with refractive index $\mu = 1.5$ has a prism angle of $30^\circ$. If a ray is incident normally on one surface,calculate its deviation. (Given: $\sin(48^\circ 36') = 0.75$)

$A$ light ray is incident at an angle of $60^o$ on a prism of angle $45^o$. When the light ray emerges from the other surface at $90^o$ to the surface,the refractive index of the material of the prism $\mu$ and the angle of deviation $\delta$ are given by: