The angle of minimum deviation for a prism is $40^o$ and the angle of the prism is $60^o$. The angle of incidence in this position will be.....$^o$

The angle of minimum deviation produced by a thin prism in air is $\delta_1$. If it is immersed in water,the angle of minimum deviation is $\left[_{a}\mu_{g}=\frac{3}{2}, _{a}\mu_{w}=\frac{4}{3}\right]$.

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:

$A$ ray of light passing through a prism $(\mu = \sqrt{3})$ suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then,the angle of prism is ..... (in degrees).

The angle of deviation produced by a thin prism when placed in air is $\delta_1$ and that when immersed in water is $\delta_2$. The refractive indices of glass and water are $\frac{3}{2}$ and $\frac{4}{3}$ respectively. The ratio $\delta_1 : \delta_2$ is

$A$ thin prism of angle $5^{\circ}$ is placed at a distance of $10\,cm$ from an object. What is the distance of the image from the object? (Given refractive index $\mu$ of prism $= 1.5$)