Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of `lambda` such that the first minima on the screen falls at a distance D from the centre O.

Explanation

From the given figure of two slit interference arrangements, we can write
`T_(2)P=T_(2)O+OP=D+x`
and `T_(1)P=T_(1)O-OP=D-x`
`S_(1)P=sqrt((S_(1)T_(1))^(2)+(PT_(1))^(2))=sqrt(D^(2)+(D-x)^(2))`
and `S_(2)P=sqrt((S_(2)T_(2))^(2)+(PT_(2))^(2))=sqrt(D^(2)+(D-x)^(2))`
The minima will occur when `S_(2)P=S_(1)P(2n-1)(lambda)/(2)`
i.e., `[D^(2)+(D+x)^(2)]^(1//2)-[D^(2)+(D-x)^(2)]^(1//2)=(lambda)/(2)` [for first minima n=1]
If x=D
we can write `[D^(2)+4D^(2)]^(1//2)-[D^(2)+0]^(1//2)=(lambda)/(2)`
`implies [5D^(2)]^(1//2)-[D^(2)]^(1//2)=(lambda)/(2)`
`implies sqrt(5)D-D=(lambda)/(2)`
`implies D(sqrt(5)-1)=lambda//2 " or D=(lambda)/(2(sqrt(5-1))`
Putting `sqrt(5)=2.236`
`implies sqrt(5)-1=2.236-1=1.236`
`D=(lambda)/(2(1.236))=0.404 lambda`