The spacing between adjacent bright fringes on a screen from a double-slit diffraction pattern is directly proportional to the distance of the screen from the slits. The path length difference between fringes must be equal to an integer multiple of the wavelength for constructive interference. This path length difference is shown to equal the wavelength multiplied by the screen distance over the slit spacing.

1. show that the spacing Ay between adjacent bright fringes appearing on the screen a distance D
away from the slits is: Ay = D Hint: The path-length difference corresponding to adjacent bright
fringes is , and you can assume small angles so sin 20.
Solution
for constructive interference,
path difference should be integral multiple of the wavelength.
let distance of m th order bright fringe from central area be y.
in the given figure, sin(theta)=path difference/d
==>theta=path difference/d
==>tan(theta)=path difference/d
==>y/D=path difference/d
==>path difference=y*d/D
making it an integral multiple of wavelength lambda,
y*d/D=m*lambda
==>y=m*lambda*D/d
so difference between m th order and (m-1) th order bright fringe is given as
(m*lambda*D/d)-((m-1)*lambda*D/d)
=lambda*D/d
…(hence proved)