The angle of a prism is $60^o$ and its refractive index is $\sqrt{2}$. The angle of minimum deviation suffered by a ray of light in passing through it is......$^o$

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.

$A$ prism is made of glass of unknown refractive index. $A$ parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^{\circ}$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^{\circ}$. If the prism is placed in water (refractive index $1.33$),predict the new angle of minimum deviation of a parallel beam of light.

$A$ combination of two prisms,one of flint glass and another of crown glass,produces dispersion without deviation. The angle of the flint glass prism is $15^o$. The angle of the crown glass prism and the net angular dispersion will be respectively: (Given: for crown glass $\mu = 1.52, \omega = 0.02$; for flint glass $\mu = 1.65, \omega = 0.03$). (in $^o$)

In the diagram,a ray is passing through a broken prism. Find the angular deviation for the ray. (in $^{\circ}$)

For a prism of prism angle $\theta=60^{\circ}$,the refractive indices of the left half and the right half are,respectively,$n_1$ and $n_2$ $(n_2 \geq n_1)$ as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1=n_2=n=1.5$. For the case of unequal refractive indices,$n_1=n$ and $n_2=n+\Delta n$ (where $\Delta n \ll n$),the angle of emergence $e=i+\Delta e$. Which of the following statement$(s)$ is (are) correct?
$(A)$ The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
$(B)$ $\Delta e$ is proportional to $\Delta n$
$(C)$ $\Delta e$ lies between $2.0$ and $3.0$ milliradians,if $\Delta n=2.8 \times 10^{-3}$
$(D)$ $\Delta e$ lies between $1.0$ and $1.6$ milliradians,if $\Delta n=2.8 \times 10^{-3}$