Calculate the angle of minimum deviation for an equilateral triangular prism of refractive index $\sqrt{3}$. (in $^{\circ}$)

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. In order for a ray to suffer minimum deviation,it should be incident at an angle of .......$^{\circ}$.

The figures represent three cases of a ray passing through a prism of angle $A$. The case corresponding to minimum deviation is

For a prism with refractive index $\mu = 1.5$,the angle of minimum deviation is equal to the angle of incidence. Find the prism angle $A$ in degrees. (Given: $\cos 41^o = 0.75$)

$A$ prism of refractive index $n_{1}$ and another prism of refractive index $n_{2}$ are stuck together (as shown in the figure). $n_{1}$ and $n_{2}$ depend on $\lambda$, the wavelength of light, according to the relation:
${n}_{1}=1.2+\frac{10.8 \times 10^{-14}}{\lambda^{2}} \text{ and } {n}_{2}=1.45+\frac{1.8 \times 10^{-14}}{\lambda^{2}}$
The wavelength for which rays incident at any angle on the interface $BC$ pass through without bending at that interface will be $....\,nm.$