Refraction of Light Through Prism

Ordinary white light, as we know, is a superimposition of waves having wavelength throughout the visible spectrum

Prism is that optical instrument which splits this white light into its constituent wavelengths

A prism dispersing white light into its composite wavelengths or colours

This breaking up of white light inside a prism happens through refraction at the prism-air interface

Let us now understand some mathematical relations incorporating Refraction of light through a prism

We consider the principal section of a prism, $ABC$

Let two normals $N_{1}M$ and $N_{2}M$ be incident on the interfaces $AB$ and $AC$ respectively

Let a ray $PQ$ be incident on $AB$ making an incident angle $i$

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 the 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 $\delta$

$\delta$ tells, to what extent the ray PQ had bent from its original path PQOR to OS, owing to refraction by the prism

Now consider $△OQT$ Let, $\angle OQT=\alpha ,\angle OTQ=\beta$

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

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 $\delta =i-r_1+e-r_2$ $\to \delta =i+e-(r_1+r_2)$

Now let us consider quadrilateral $AQMT$ $$\begin{gathered} \angle AQM+\angle ATM=180^0 \\ \text{(MQ and MT are normals)} \end{gathered}$$

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

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

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

As $A=r_1+r_2$ and $\delta =i+e-(r_1+r_2)$ $\therefore \delta =i+e-A$ $\to \delta +A=i+e$

Thus we get a relation between the angle of deviation $\delta$, the angle of prism $A$, the angle of incidence $i$ and the angle of emergence $e$

Revision

The relation between $\delta$ and $A$ is $\delta +A=i+e$ also $A$ can be written as $A=r_1+r_2$

The relation $\delta +A=i+e$relates between the angle of deviation $\delta$, the angle of prism $A$, the angle of incidence $i$ and the angle of emergence $e$

The End