$A$ ray of monochromatic light is passing through an equilateral prism $(ABC)$ as shown in the figure. The refracted ray $(QR)$ is parallel to its base $(BC)$ and the angle of incidence $(i)$ is $50^\circ$. Then the angle of deviation $(\delta)$ is: (in $^\circ$)

$A$ given ray of light suffers minimum deviation in an equilateral prism $P$. Additional prisms $Q$ and $R$ of identical shape and material are now added to $P$ as shown in the figure. The ray will suffer

If the angle of a prism is $60^{\circ}$ and the angle of minimum deviation is $40^{\circ}$,then the angle of refraction will be (in $^{\circ}$)

One face of a prism with a refracting angle of $30^{\circ}$ is coated with silver. $A$ ray incident on the other face at an angle of $45^{\circ}$ is refracted,reflected from the silver-coated face,and retraces its path. The refractive index of the prism is:

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:

For a prism with a prism angle of $60^\circ$ and a refractive index of $\mu = \sqrt{2}$,the angle of minimum deviation is .......... $^\circ$.