The cross-section of a prism $(\mu = 1.5)$ is an equilateral triangle. $A$ ray of light is incident perpendicular on one of the faces. The angle of deviation of the ray is......$^o$

The refractive index of the material of a small-angled prism is $1.6$. If the angle of minimum deviation is $4.2^{\circ}$, the angle of the prism is: (in $^{\circ}$)

$A$ ray is incident at an angle of incidence $i$ on one surface of a small angle prism (with angle of prism $A$) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$,then the angle of incidence is nearly equal to

The angle of minimum deviation is equal to the angle of the equilateral prism. At what angle of incidence $i$ will this minimum deviation occur?

$A$ 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 $....\,nm.$

In the given figure,the face $AC$ of the equilateral prism is immersed in a liquid of refractive index $n$. For an incident angle of $60^{\circ}$ at the side $AC$,the refracted light beam just grazes along the face $AC$. The refractive index of the liquid is $n = \frac{\sqrt{x}}{4}$. The value of $x$ is (Given refractive index of glass $= 1.5$)