The fringe width in Young’s double slit experiment increases when

Explanation

Fringe Width. It is the distance between any two successive fringes in interference pattern.
Let `S_(1), S_(2)` be the two fine slits illuminated by a monochromatic sources of wavelength `lambda`.
Intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between `S_(2)P` and `S_(1)P`.

From `S_(1)` draw `S_(1)A` perpendicular to `S_(2)P`. Since `/_S_(2)S_(1)A to theta` [D is very large as compared to d], therefore, `/_S_(1)AS_(2) to 90^(@)` and `AP ~~ S_(1)P`.
`:. "Path difference" = S_(2)P-S_(1)P = (S_(2)A + AP) - S_(1)P` or `"Path difference" = S_(2)A = d sin theta`
Since `theta` is very small, `sin theta` can be replaced by `tan theta` .
`:. "Path difference" = d tan theta = d(y)/(D)`
Constructive interference [Bright Fringes). For bright fringes, the path difference should be equal to an even multiple of `lambda // 2`.
`:. (dy)/(D)=2n(lambda)/(2)=n lambda`
or `y=n(lambda D)/(d)`
If n = 0, `y_(0)=0`, which is the position of central maxima.
If n = 1, `y_(1)=(D)/(d)lambda`, which is the position of first maxima.
If n = 2, `y_(2)=(2D)/(d)lambda`, which is the position of
Similarly `y_(n-1)=(n-1)(D)/(d)lambda`
and `y_(n)=n(D)/(d)lambda`
Fringe width `beta=y_(n)-y(n-1)=(D)/(d)lambda[n-(n-1)]`
`beta=(D)/(d)lambda` ...(i)
Destructive interference [Dark Fringes]. For dark fringes, the path difference should be an old multiple of `1 // 2`.
`(d)/(D)y=(2n-1)(lambda)/(2)`
or `y=(D)/(d)(2n-1)(lambda)/(2)`
If n = 1, `y_(1)=(D)/(d)(lambda)/(2)`, which is the position of 1st minima.
If n = 2, `y_(1)=(D)/(d)(2lambda)/(2)`, which is the position of 2nd minima.
If n = 3, `y_(1)=(D)/(d)(3lambda)/(2)`, which is the position of 3rd minima.
`:.` Fringe width, `beta=y_(3)-y_(2)=(D)/(d)lambda`
or `beta=(D)/(d)lambda` ...(ii)
From (i) and (ii), we conclude that bright and dark fringes have equal fringe width.
`beta prop D`
`prop lambda`
`prop (1)/(d)`
If d = 0, `beta= infty` i.e. dark and bright bands will be infinitely well spaced and there will be uniform illumination.