Double slit diffraction maxima

[HorseBox]

What effect does the distance between the two slits have on the intensity of the maxima?

 

[Andy Resnick]

It can have a large effect, if I understand what you mean by 'intensity'. What have you come up with so far?

 

[HorseBox]

I just started learning the concepts yesterday what I've got so far is that if I have a screen with 2 slits each of a width equal to or smaller than the wavelength of the light then the slits can be considered single point sources of waves. If I place another screen far away from the screen with the slits a diffraction pattern will form on it consisting of fringes of light (maxima) and dark strips (minima) in between them. If I'm not mistaken it can be determined whether any point on the screen is a maxima or minima by comparing the length L of the paths taken by a wave traveling from each of the slits. If there is a difference equal to any multiple of the wavelength of the light waves then it will be a maxima because it will be pure constructive interference and the opposite being true for path length differences of any multiple of the wavelength plus a half wavelength. The equation I got is
\frac{L_1 - L_2}{d} = sin\theta
I just read that this equation is the same as
\frac{L_1 - L_2}{d} = m\lambda
where m is determines the amount of destructive interference at the point on the screen. I can see now that since the wavelength remains constant altering the distance between the slits d will alter the value of m so will effect the amount of destructive interference between the two waves at any point.

EDIT: Then again point on the screen directly opposite the space between the 2 slits will always be a maximum because the waves from each slit always travel the same distance to reach it.