Prism formula and Related Terms

We must have seen sunlight coming from the sun.

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 $\delta$.

Let's now derive the prism formula.

Consider quadrilateral $AQMT$ $\angle AQM+\angle ATM=180^0$.

So, we can say quadrilateral $AQMT$ is cyclic $\therefore \angle A+\angle M=180^0\to \text{Eqn 1}$

In $△MQT$ $\angle r_1+\angle r_2+\angle M=180^0\to \text{Eqn 2}$

From Eqn 1 and 2 we get, $\angle r_1+\angle r_2+\angle M=\angle M+\angle A$ $\therefore \angle r_1+\angle r_2=\angle A$

Consider $△OQT$ Let, $\angle OQT=\alpha ,\angle OTQ=\beta$ ,

$$\begin{gathered} \therefore \\ \delta =\alpha +\beta \\ \text{(Sum of interior angles=Ext angle)} \end{gathered}$$

Now from the diagram, we can write the interior angles as, $\alpha +r_1=i$ $\therefore \alpha =i-r_1$

Similarly $\beta =e-r_2$

Now simplifying the equation $\delta =\alpha +\beta$ we get $\delta =i-r_1+e-r_2$ $\to \delta =i+e-(r_1+r_2)$.

As $A=r_1+r_2$ and $\delta =i+e-(r_1+r_2)$ $\therefore \delta =i+e-A$ $\to \delta +A=i+e\to \text{Eqn 3}$

At minimum deviation, the angle of incidence is equal to the angle of emergence $\therefore \delta ={\delta }_{m}wheni=e$

Again from the diagram and the symmetry we can write that the angle refractions are also the same $\therefore i=e,r_1=r_2$

Now from equation $3$ we get, $\delta +A=i+e$ $\therefore {\delta }_m+A=i+i$ $\to i=\frac{{\delta }_m+A}{2}$

Again $A=r_1+r_2$ and when $\delta ={\delta }_m$ $\therefore A=r+r$ $\to r=\frac{A}{2}$

From Snell's law, we can write refractive index ,$\mu =\frac{\sin \mathbf{i}}{\sin \mathbf{r}}$ $\mu =\frac{\sin (\frac{A+{\delta }_m}{2})}{\sin (\frac{A}{2})}$

The equation $\mu =\frac{\sin (\frac{A+{\delta }_m}{2})}{\sin (\frac{A}{2})}$ 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 $\delta$ and $A$ is $\therefore \delta +A=i+e$ also $A$ can be written as $A=r_1+r_2$

In case of minimum deviation $\delta ={\delta }_m$, $i=e$. Also $i$ can be written as $i=\frac{A+{\delta }_m}{2}$

The prism formula is $\mu =\frac{\sin (\frac{A+{\delta }_m}{2})}{\sin (\frac{A}{2})}$

The End