Fringe Pattern Brightness for Young's Double-slit experiment Perhaps I have missed something in my notes, but I have noticed when looking at different sources that some textbooks/sites state that the fringe brightness for the young's experiment is the same for all the bright fringes. Others, say that the brightness "falls off" with angle theta. Well, which is it? And how do you calculate it, not using derivatives if possible... I'm preparing for the MCAT exam and calculus is not tested. Thanks!
Here are some pictures that I've come across either in different textbooks or websites:





 A: The Intenisty distribution of a double slit diffraction is 
$$I=I_0\cos^2[\frac{\pi d \sin(\theta)}{\lambda}]$$
Where we define $\tan(\theta)=\frac{\Delta y}{L}$.  Remember that the y spacing is proportional to the slit seperation so we end up that 
$$\tan\theta\approx \frac{d}{L}$$
In the one figure, it is stated that L>>d, so we use the approximation that $sin\theta=\theta$, so that equation for I above would have no angular dependance in that case.  
In general there will be an angular dependance.
A: I'm sorry for the (extremely) delayed response, but the other answer fails to answer the question.
The reason for the intensity falling off with distance is mainly because of the slit width itself. Because of this width, the individual Fraunhoffer diffraction pattern is superimposed in the final observed pattern.
If the slit width were to be infinitesimally small, then you'd observe the ideal $\cos^2(\theta)$ graph. But since the slit itself has a finite width, it also contributes a factor, which is the sinc$^2$($\theta$) function.
It is this sinc$^2$($\theta$) function that causes the fall of intensity with distance. Basically, what you observe in general is a combination of the one-slit diffraction pattern, and the 2 slit interference of YDSE.
