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.

Options

A. `(lambda)/(2.472)`
B. `(lambda)/(2.236)`
C. `(lambda)/(1.227)`
D. `(lambda)/(3.412)`

Answer

A

Explanation

From diagram
`T_(1)P=T_(1)O-OP=(D-x)`
`T_(2)P=T_(2)O-OP=(D+x)`
Now `S_(1)P=sqrt((S_(1)T_(1))^(2)+(T_(1)P)^(2))=sqrt(D^(2)+(D-x)^(2))`
`S_(2)P=sqrt((S_(2)T_(2))^(2)+(T_(2)P)^(2))=sqrt(D^(2)+(D+x)^(2))`
Path difference `S_(2)P-S_(1)P=(lambda)/(2)`, for first minimum to occur
`sqrt(D^(2)+(D+x)^(2))-sqrt(D^(2)+(D-x)^(2))=(lambda)/(2)`
The first minimum falls at a distance D from the center, i.e., `x=D.`
`[D^(2)+4D^(2)]^(1//2)-D=(lambda)/(2)`
`D(sqrt(5)-1)=(lambda)/(2)`
`D(2.236-1)=(lambda)/(2)" , "D=(lambda)/(2.472)`