Discuss the phenomenon of refraction through a prism. Prove that `delta = (mu -1 ) A` where the symbols have their usual meaning.

Explanation

Refraction. The phenomenon of change in path of a ray of light as it passes from one medium into another is called refraction.
Let a ray DE strikes the face AB of the prism at an angle of incidence i and emerges out at an angle e. Let `r_(1)` and `r_(2)` be the angles of refraction and `delta` be exterior angle of deviation.
`delta` is exterior angle of triangle HEF
`:. delta=1+2=(i-r_(1))+(e-r_(2))`
or `delta=i+e-(r_(1)+r_(2))` ...(1)
In quadrilateral AEOF, we have
`/_ AEO + A +/_AFO +/_O=360^(@)`
or `90^(@)+A+90^(@)+ /_O=360^(@)`
or `A+ /_O=180^(@)`

Also, in triangle FEO,
`r_(1)+r_(2)+ /_O=180^(@)`
`A+ /_O=r_(1)+r_(2)+ /_O`
or `A=r_(1)+r_(2)` ...(ii)
Substituting in equation (i), we have
`delta=i+e-A`
or `A+delta=i+e`
For small angled prism,
At face AB,
`(sin i)/(sin r_(1))=(i)/(r_(1))=(mu)`
or `i=mu r_(1)`
Similarly, at face AC,
`e=mu r_(2)`
Putting the value of i and e in equation (iii), we get
`delta=mu r_(1)+ mu r_(2)-A=mu(r_(1)+r_(2))-A`
Using equation (ii), we get
`delta=(mu-1)A` ...(iv)
If we repeat the experiment with different angles of incidence, then we get a graph as shown in the figure when i is increased, `delta` decreases first, reaches a minimum and starts increasing again. For a particular value of `delta`, there are two angles of incidence `i_(1)` and `i_(2)`, At minimum deviation,

`delta=delta_(m)` and `i_(1)=i_(2)=i`
Hence `r_(1)=r_(2)=r`
`:. r+r=2A` or `r=(A)/(2)`
And `delta_(m)=i+i-A`
or `2i=A+delta_(m)`
or `i=(A+delta_(m))/(2)`
Now `mu=(sin i)/(sin r)=(sin(A+delta_(m))/(2))/(sin (A)/(2))`