If $r_1$ and $r_2$ are the angles of refraction at the first and second faces of a prism,respectively,then the angle of the prism is:

For a thin prism,$\delta_1$ is the angle of deviation produced when the prism is placed in air. When the prism is immersed in water,the angle of deviation produced is $\delta_2$. Given ${ }_{a} \mu_{g}=\frac{3}{2}$ and ${ }_{a} \mu_{w}=\frac{4}{3}$. The ratio $\delta_2: \delta_1$ is

$A$ 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: $\sin(48^\circ 36') = 0.75$)

The refractive index $(R.I.)$ of a prism is $\sqrt{\frac{7}{3}}$ and the angle of the prism is $60^{\circ}$. The limiting angle of incidence of a ray that will be transmitted through the prism is .....$^{\circ}$

$A$ ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to $\frac{3}{4}$ of the angle of the prism. The angle of deviation is (in $^{\circ}$)

As shown in the diagram, when the incident ray is parallel to the base of the prism, the emergent ray grazes along the second surface. If the refractive index of the material of the prism is $\sqrt{2}$, the angle $\theta$ of the prism is: (in $^{\circ}$)