Calculating total number of fringes in Young's Double Slit Experiment I know that the $n^{th}$ order maxima can be given by $\frac{n\lambda D}{d}$. But theoretically n can go up to infinity and the intensity will go down. However, is there any way to approximate the visible bright spots that can be observed in experiments?
 A: I assume that you want to approximate the number of observable maxima. For very narrow slits the number of maxima is $n <= d/\lambda$ on each side of the central maximum. So the number of maxima is 2n+1. These are all observable. The individual slits also produce $m <=w/\lambda$ maxima. For a narrow slit m=0. The total number of maxima is then 2n+2m+1. This formula does not take into account that maxima may be merged.
Of course this is what you can observe under very good conditions. Intense monochromatic light, a perfect slit, low background, and a good detector will be required if you want to see them all. 
A: In Young's slits, the two beams that interfere have a width limited by the diffraction by the slits. The fringes are visible only in the common part of the two beams.
By neglecting the distance between the slits, the angular width associated with the diffraction is $2(\lambda /a)$and the angular width of a fringe is $\lambda /d$
As the central fringe is bright, we will roughly have $N=1+2d/a$ visible fringes.
It is an approaching reasoning that may forget certain elements!
Sorry for my poor english !
A: 
This is variation in number of fringe widths with wavelength
