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

Obtain the relation between incidence angle,emergence angle,prism angle,and deviation angle for refraction through a prism.

$A$ small-angled prism is made of a material of refractive index $\frac{3}{2}$. The ratio of the angles of minimum deviation when the prism is placed in air and in water of refractive index $\frac{4}{3}$ is:

The path of a refracted ray of light in a prism is parallel to the base of the prism only when the

$A$ 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 $^{\circ}$)

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: