The color of the sunlight looks white but it is made up of several colors.

Prism can split white light into its constituent colors.

Light rays passing through a prism deviates from its original path due to refraction.

Owing to this refraction, the prism disperses white light into a full spectrum of it's constituents.

In order to understand these prisms, let us first understand some terms.

Consider a prism with 2 triangular and 3 rectangular sides

Now a cross-section of the prism, would look something as shown above, $△ABC$

Let's draw two normals $N_{1}M$ and $N_{2}M$ on $AB$ and $AC$ respectively.

Now let a ray PQ be incident on $AB$ at a point $Q$ such that the angle of incidence is $i$ (as shown),

As the ray is travelling from rarer to denser medium; it bends towards the normal.

Then it refracts at $AC$ and bends away from the normal (denser to rarer) forming an emergent ray.

This emergent ray forms an angle $e$ with the normal at $AC$ known as the angle of emergence.

The angles of refraction are $r_{1}$ and $r_{2}$ respectively.

Now if the emergent ray is extended backwards to $O$, then we get the angle of deviation, denoted by $δ$.

Let's now derive the prism formula.

Consider quadrilateral $AQMT$$∠AQM+∠ATM=180_{0}$.

So, we can say quadrilateral $AQMT$ is cyclic $∴∠A+∠M=180_{0}→Eqn 1$

In $△MQT$$∠r_{1}+∠r_{2}+∠M=180_{0}→Eqn 2$

From Eqn 1 and 2 we get, $∠r_{1}+∠r_{2}+∠M=∠M+∠A$$∴∠r_{1}+∠r_{2}=∠A$

Consider $△OQT$ Let, $∠OQT=α,∠OTQ=β$ ,

$∴δ=α+β(Sum of interior angles=Ext angle)$

Now from the diagram, we can write the interior angles as, $α+r_{1}=i$$∴α=i−r_{1}$

Similarly $β=e−r_{2}$

Now simplifying the equation $δ=α+β$ we get
$δ=i−r_{1}+e−r_{2}$$→δ=i+e−(r_{1}+r_{2})$.

As $A=r_{1}+r_{2}$ and $δ=i+e−(r_{1}+r_{2})$$∴δ=i+e−A$$→δ+A=i+e→Eqn 3$

At minimum deviation, the angle of incidence is equal to the angle of emergence $∴δ=δ_{m}wheni=e$

Again from the diagram and the symmetry we can write that the angle refractions are also the same $∴i=e,r_{1}=r_{2}$

Now from equation $3$ we get, $δ+A=i+e$$∴δ_{m}+A=i+i$$→i=2δ_{m}+A $

Again $A=r_{1}+r_{2}$ and when $δ=δ_{m}$$∴A=r+r$$→r=2A $

From Snell's law, we can write refractive index ,$μ=sinrsini $$μ=sin(2A )sin(2A+δ_{m} ) $

The equation $μ=sin(2A )sin(2A+δ_{m} ) $ gives the relation between the refractive index of the material of the prism, the minimum deviation and the angle of the prism

Revision

The relation between $δ$ and $A$ is $∴δ+A=i+e$
also $A$ can be written as $A=r_{1}+r_{2}$

In case of minimum deviation $δ=δ_{m}$, $i=e$. Also $i$ can be written as $i=2A+δ_{m} $