Light interference single particle 4 slits experiment - what are interpretations apart from many-worlds interpretation? Someone knowledgeable about physics advised me to read a book "David Deutsch: The Fabric of Reality, 1997". The book gives experiment like two-slit one, difference is that 2-slits compared to 4-slits results for single photons. [experiment: red light laser, 1st part: 2 slits are said to be one-fifth of a millimetre apart, 2nd part - one slit is added in between and one to the side of 1st two, resulting in one-tenth of millimeter distance, width of slit is not given, overall experiment is not described in much detail] -> [result: the pattern changes to less number of light vertical "bars"]. 
In Internet I found about 2-slits experiments, e.g. http://www.youtube.com/watch?v=GzbKb59my3Ua and wikipedia http://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles.
David gives many-worlds explanation by introducing shadow photons from parallel universes and do not mention particle-wave thing. 
But, visible light wave is around 1000 nanometers, so distance between slits is more than 100 wave lengths of light. Maybe amplitude of photon is 100 times more than wave length and it's path is bent by several slits together? Otherwise how light bar disappeared due to adding more slits apart from many-worlds interpretation (MWI)?
 A: The disappearing interference lines can also be well understood as typical 4-slit optical interference.  
Let $D$ be the distance from the slits to the screen and $d$ the separation between 2 adjacent slits. The intensity pattern on the screen is then a function $I(x)$ of the on-screen position $x$ as shown in the figure below:

To see why some lines go missing, let's start with the general interference formula for N equidistant slits and light of wavelength $\lambda$ (see Sec.9.2 here for derivation and eq.(19) therein):
$$
\frac{I_N(x)}{I_N(0)} = \left( \frac{\sin\left( N \frac{\pi d}{\lambda}\frac{x}{D}\right)}{N\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2
$$
For $N=2$ we get the regular 2-slit interference pattern,
$$
\frac{I_2(x)}{I_2(0)} = \left( \frac{\sin\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)}{2\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2 =  \left( \frac{2 \sin\left(  \frac{\pi d}{\lambda}\frac{x}{D} \right) \cos\left(\frac{\pi d}{\lambda}\frac{x}{D}\right)}{2\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2 = \cos^2\left(\frac{\pi d}{\lambda}\frac{x}{D}\right)
$$
Let's see what we get for $N=4$:
$$
\frac{I_4(x)}{I_4(0)} = \left( \frac{\sin\left( 4 \frac{\pi d}{\lambda}\frac{x}{D}\right)}{4\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2\\ = \left( \frac{2\sin\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)}{4\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2  \\
= \left( \frac{4\sin\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)}{4\sin\left(\frac{\pi d}{\lambda}\frac{x}{D} \right)}\right)^2 \\= \left( \cos\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)\right)^2\\
$$
or 
$$
\frac{I_4(x)}{I_4(0)} = \cos^2\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos^2\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)
$$
We can see that the 2nd factor above is a 2-slit pattern for a pair of slits a distance $2d$ apart. At this point we may conclude that the missing lines, compared to a (2d)-pattern, are due to the modulating 1st factor, which is itself a 2-slit pattern. A plot of the 2 patterns side by side looks as follows, clearly showing "missing lines":

However, there is one other, perhaps more physical way to interpret this result. 
Let us rewrite the last expression in a slightly more revealing form by transforming the cosine product into a sum ($\;\cos(\alpha)\cos(\beta) = \frac{1}{2}\left[\cos(\alpha + \beta) + \cos(\alpha - \beta)\right]\;$):
$$
\frac{I_4(x)}{I_4(0)} = \left( \cos\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( 2 \frac{\pi d}{\lambda}\frac{x}{D}\right)\right)^2
\sim \left( \cos\left( 3\frac{\pi d}{\lambda}\frac{x}{D}\right) + \cos\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\right)^2  \\
=\cos^2\left( \frac{\pi d}{\lambda}\frac{x}{D}\right) +  \cos^2\left( 3\frac{\pi d}{\lambda}\frac{x}{D}\right) + 2 \cos\left( \frac{\pi d}{\lambda}\frac{x}{D}\right)\cos\left( 3 \frac{\pi d}{\lambda}\frac{x}{D}\right)
$$
We recognize again the first term as a 2-slit pattern for separation d. The 2nd term is also a 2-slit pattern, but for separation $3d$. Given the 4-slit geometry, we can easily identify the origins of these contributions: 


*

*The 1st term is the interference pattern produced by the 2 inner slits, with separation d.

*The 2nd term is the corresponding pattern produced by the 2 outer slits, with separation 3d.

*The 3rd term obviously accounts for the interference of the interference patterns themselves. We may call this last contribution a second order interference.


If we now compare the overall 4-slit pattern to the patterns produced by the outer and inner pairs of slits, we obtain something like the following: 

Again we see "missing lines". But since the first 2 terms do not subtract, but actually add two independent 2-slit interference patterns, it must be that the missing peaks are due to the 3rd term: 
The pattern produced by the outer pair of slits interferes destructively with that produced by the inner pair and wipes out non-overlapping 2-slit maxima. 
As always with optical interference, all that is involved is a superposition of electromagnetic fields.      
A: The answer by @urdv  gives the mathematics of destructive interference in the classical framework.
The classical framework emerges from the quantum mechanical framework smoothly, the individual photons that build up the classical wave follow in bulk the classical Maxwell equation solutions of classical electromagnetic fields because their wavefunctions are the solutions of a form of  quantized Maxwell equations. Thus the classical result holds  for the photons impinging on slits too.
Please keep in mind that the many worlds interpretation is just that, an interpretation of the mathematics involved in quantum mechanics, assigning mathematical probabilities to "real" hypothetical worlds. The example you are discussing by construction  of the many worlds, cannot be in conflict with the quantum mechanical solutions.
