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Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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It will be easiest to describe this using an x-y coordinate system.

Suppos the two sources are on the y axis, one 3 meters above the origin and the other 3 meters below it.

The segment between the two sources is 6 meters long, and is part of the y axis, running from 3 meters above to 3 meters below the origin.

The origin is the midpoint of this segment. The perpendicular bisector of this segment is the line perpendicular to it, passing through its midpoint.

The line perpendicular to this segment which passes through the midpoint, which is the origin, is the x axis.

Imagine two rays, one from each source, both directed at some distant point not on the x axis. For convenience let's assume that a line from the origin to the point makes a 30 degree angle with the positive x axis, as measured counterclockwise from the axis. You should sketch this.

Since the point is distant, the rays from the two sources will be very nearly parallel to this line, so both will make angles of very nearly 30 degrees with the x axis. You should sketch these lines; you won't be able to sketch them all the way to the distant point, but sketch them until they reach the edge of your paper. The distance point is supposed to be really distant, so the lines won't get significantly closer to one another by the time they reach the edge.

Now from the top source, sketch a projetion line to the source to the lower ray. Remember that a projection line is perpendicular to the line on which it projects, so your projection line will be perpendicular to the lower ray.

The line segment between the sources, your projection line, and the short part of the lower ray from the lower source to the projection line form a right triangle. The angle at the top of that triangle is the same as the angle made by the ray with the x axis; if everything is drawn with impossibly complete accuracy that angle will be 30 degrees.

You can check out the figure at

http://www.google.com/imgres?imgurl=http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/doubsli.gif&imgrefurl=http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html&h=371&w=601&sz=32&tbnid=zI5bEdeTvpf_mM:&tbnh=76&tbnw=123&zoom=1&usg=__g1ZilwYkhv-Gb2O2o3yRZ_bGA34=&docid=wXCOoz_70y-iBM&sa=X&ei=2cpwUa_bMIHc8wSglIHgBA&ved=0CEUQ9QEwAg&dur=173

or just google 'two slit interference' and look for similar images. In the image I referenced, the line D would correspond to the x axis, the distance d would be 6 meters, the distance 'delta' (the Greek-looking letter on the short side of the triangle) is the extra distance traveled by the lower ray, and the 'distant point' is included in the picture so the upper and lower rays are not quite parallel.

Along either the upper or lower ray, the distance from the projection line to the distance point is the same. The extra distance traveled by the lower ray is the short side of the triangle.

If that extra distance is half a wavelength, then assuming that the sources are in phase, a peak emitted from the upper ray, upon raching the distant point, will meet a valley emitted by the lower ray and the two will cancel. The same will be true if the extra distance is one wavelength plus half a wavelength, or two wavelengths plus half a wavelength, etc..

If the extra distance is a whole wavelength then peaks will meet peaks at the distance point and the waves will reinforce.

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There is also a very neat simulation at

http://science.sbcc.edu/physics/flash/2%20slit%20interference.html

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