The minimum value of d os that there is a dark fringe at O is `d_(min)`. For the value of `d_(min)`, the distance at which the next bright fringe is formed is x. Then

Options

A. `d_(min) = sqrt(lambda D)`
B. `d_(min) = sqrt((lambda D)/(2))`
C. `x = (d_(min))/(2)`
D. `x = d_(min)`

Answer

B, D

Explanation

There is a dark fringe at 'O' is the path difference
`Delta x = ABO - AO'O= lambda//2`
`Delta x = 2sqrt(D^(2)+d^(2)) - 2D = 2D sqrt(1+ (d^(2))/(D^(2))) - 2D`
`Delta x = 2D [(1+(d^(2))/(D^(2)))^(1//2)-1] = 2D [1+ (d^(2))/(2D^(2))-1]`
`Delta x = 2Dxx(d^(2))/(2D^(2))= (d^(2))/(D) = (d^(2))/(D) = lambda//2 implies d^(2) = (lambda D)/(2)`
`implies d_(min) = sqrt((lambda D)/(2)`
There bright fringe is fromed at P if the path difference
`Delta x^(1) = AO'P - ABP = lambda`
`Delta x^(1) = D + sqrt(D^(2)+x^(2)) - sqrt(D^(2) + d^(2)) - sqrt(D^(2)+(x-d)^(2)) = lambda`
`= (x^(2))/(2D) - (d^(2))/(2D) - ((x^(2)+d^(2))-2xd)/(2D) = lambda`
Given `d = d_(min) then (-d^(2))/(D) + (xd)/(D) = lambda` on solving