I'm having trouble understanding the question as well as the explanation.
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Just to copy the question, it says: "If a helium-neon laser is used as the light source (λ = 6328 Å), how far should the movable mirror be displaced to move from one bright fringe to the next one?"
First of all, as a strategic point, don't let the fact that there are Greek letters and strange units/Swedish letters throw you off. Think logically about the experiment they're doing.
First of all, why do we use the helium-neon laser? Apparently because it gives one wavelength, i.e., is monochromatic. Also, what are these "bright fringes"? According to the passage, a bright fringe is a "maxima" (paragraph 4). But again, "maxima" of what? In this case, the "maxima" are distances that give us constructive interference. And then we ask again -- constructive interference? Of what?
Constructive interference (and any type of interference, for that matter) requires two waves to interact. For the interference to be judged as "constructive," the two waves need to add together when they are in phase. This occurs when the two waves' electric and magnetic fields are oscillating with the same wavelength, and when they are hitting their peaks and troughs at the same time.
The two waves in this passage are created by the beamsplitter. The beamsplitter allows half of the waves to pass through and travel to mirror M2; another half is reflected and goes to mirror M1. Both waves reflect off of M2 and M1, respectively, and then travel back to the beamsplitter. They are all directed to the detector at that point.
When the distances d1 and d2 are equal, you can imagine that the two waves are traveling the exact same distance -- however far it is from the radiation source to the beam splitter plus two times the distance to the mirror M1 or M2 (remember, it has to travel there and also come back), and then the distance from the splitter to the detector. The only part varying here is that distance to the mirror M1 or M2, so let's just focus on that. In this first case, if d1 = d2, then each wave has traveled 2d for its distance. How do we get two waves to constructively interfere? Again, they have to be in phase. If we start pushing out d2, when will the two waves be in sync again? Well, when one wave is a full wavelength "ahead" of the other -- their peaks and troughs will still line up, even if one is one wavelength "behind". But remember that the wave actually has to travel double the distance we push out the mirror, since it will travel this new distance to the mirror as well as away from the mirror and back towards the beamsplitter.
In this example, the wavelength is 6328 Å. To add that distance to one of the waves, we'd have to push its mirror back half that distance (remember the doubling), or 3164 Å. Then the two waves will be in sync again, and constructively interfere. This creates the next bright fringe.
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