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Can you create a young's fringe pattern using two lasers of the same type, with fairly divergent beams, in place of the single source and two slits, commonly used to create this type of pattern? How?

2006-12-11 23:10:12 · 5 answers · asked by himself202 1 in Science & Mathematics Physics

5 answers

It seems very unlikely to me that this is possible.
Creating a fringe pattern requires two coherent light sources (=the phase of the light from one source can be predicted from the phase from the light of the other source)
I don not see how two separate lasers can be made coherent.

2006-12-11 23:22:54 · answer #1 · answered by mitch_online_nl 3 · 1 0

You can certainly get interference from two separate lasers operating at the same frequency, but this would not be the same as the Youngs slit experiment. Youngs slits is defineed to sample two points on a single wavefront - by implication this means a single source (a light bulb will do just fine - it does not have to be especially coherent).

2006-12-12 07:42:18 · answer #2 · answered by Anonymous · 1 0

No. For the fringes to appear, you need the 2 sources to be coherent between them. You can't get that with 2 lasers. That's why every set-up using interference finds a way to split the laser first so you have "2" coherent sources.

2006-12-12 07:26:14 · answer #3 · answered by Vincent L 3 · 1 0

Yes....There is a company called Banner. They manufacture fiber optics for the automation/robotics industry used in place of solid wires to convey information to inputs and outputs for various electrical controls and PLC's. When light from the fiber optics converge to form a triggering convergence beam to signal some other input or output, they make the darndest patterns. Could this be called a younger fringe pattern???
P.S. I just had to have fun with this one...

2006-12-12 07:23:05 · answer #4 · answered by dewhatulike 5 · 0 1

u only have to change the optical path of one of them. U can also make it with only one laser an using a beam spliter.

2006-12-12 07:28:21 · answer #5 · answered by g 2 · 0 1

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