double-slit-experimentwaves
I was taught in class that the double slit diffraction pattern would always have bright fringes of same length. We derived the formula: $y=m(\lambda)(L)/d$, which also shows this equal distance.
I look at these pictures now, and I see a central maximum with almost twice the largeness of other fringes. What is going on?
In this case whoever set the question has chosen wavelengths with a nice ratio and not too far apart, so the first coincidence is at $n_{600}$ = 3 and $n_{450} = 4$. However you're quite correct that this isn't a general rule. For example if the two wavelengths, $\lambda_1$ and $\lambda_2$ were both primes they would coincide when $n_1 = \lambda_2$ and $n_2$ = $\lambda_1$.
The experiment looks like:
You can calculate the path lengths $A$ and $B$ using Pythagoras' theorem. The 100th bright fringe is positioned where the different in path length is 100$\lambda$ i.e.
$$ B - A = 100\lambda$$
The resulting equation is messy but easily solved numerically. I got $s \approx 1.51$m at the 100th fringe, so the spacing between the 100th and -100th fringes would be about $3.02$m.
Best Answer
I think maybe it's just a bad photo. The double slit diffraction pattern is a convolution of the pattern from a pair of delta functions with that of a single slit. i.e. You have the regularly spaced double slit pattern, but this is modulated by the pattern produced by a single slit of width equal to that of the slits used for your double slit experiment. In the photo you show it looks like maybe poor resolution/focus/contrast has lost the distinction of the finer double slit fringes?
The photo below from the wikipedia page on "Introduction to quantum mechanics" shows this more clearly.
Or maybe, as others have said, it is just a photo of a single slit diffraction pattern put there in error