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. (a) `sqrt((3lambdaD)/(2))`
B. (b) `sqrt(lambdaD)`
C. (c) `sqrt((lambdaD)/(2))`
D. (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`=Path difference between the waves after emerging from slits.
`Delta_1=SS_1-SS_2=sqrt(D^2+d^2)-D`
and `Delta_2=S_1O-S_2O=sqrt(D^2+d^2)-D`
`:. Delta=2{(D^2+d^2)^(1/2)-D}=2{(D^2+(d^2)/(2D))-D}`
`=d^2/D` (From Binomial expansion)
For obtaining dark at O, `Delta` must be equals to
`(2n-1)lambda/2`, i.e., `d^2/D=(2n-1)lambda/2impliesd=sqrt(((2n-1)lambdaD)/(2))`
For minimum distance `n=1` so `d=sqrt((lambdaD)/(2))`.