Two ideal slits `S_(1)` and `S_(2)` are at a distance `d` apart, and illuninated by light of wavelength `lambda` passing through an ideal source slit `S` placed on the line through `S_(2)` as shown. The distance between the planes of slits and the source slit is `D.A` screen is held at a distance `D` from the plane of the slits. The minimum value of `d` for which there is darkness at `O` is `(dlt lt D)`

Options

A. `sqrt((3lambdaD)/2)`
B. `sqrt(lambdaD)`
C. `sqrt((lambdaD)/2)`
D. `sqrt(3lambdaD)`

Answer

C

Explanation

Path difference between the waves reaching at P,
`Delta=Delta_(1)+Delta_(2)`
where, `Delta_(1)` initial path difference
`Delta_(2)` = difference between the waves after emerging from slits
Now, `Delta_(1)=SS_(1)-SS_(2)=sqrt((D^(2)+d^(2)))-D`
and `Delta_(2)=S_(1)O-S_(2)O=sqrt((D^(2)+d^(2)))-D`
`:.Delta=2{sqrt((D^(2)+d^(2)))-D}=2{(D^(2)+d^(2))^(1//2)-D}`
`=2{(D+d^(2)/(2D))-D}` from binomial expansion
`=d^(2)/D`
For obtaining dark at O, `Delta` must be equals to `(2n-1)lambda/2`
i.e., `d^(2)/D=(2n-1)lambda/2`
`:. d^(2)=((2n-1)lambdaD)/2 or d=sqrt(((2n-1)lambdaD)/2)`
For minimum distance n=1
So, `d=sqrt((lambdaD)/2)`