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PHY 202
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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Test 2 Practice Questions
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Here is a test question I found on the Testing page while practicing/studying (http://vhmthphy.vhcc.edu/tests/ph2/test2c.htm):
Two sources separated by 5.85 meters emit waves with wavelength .61 meters emit waves in phase. The waves travel at identical velocities to a distant observer. At any point along the perpendicular bisector of the line segment connecting the two points, the two waves will arrive in phase an hence reinforce. At what nonzero angle with the perpendicular bisector will the first interference minimum be observed?
I struggled with this problem somewhat and wanted to see if my reasoning was correct (I take the exam on 3/19 at 9AM).
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The first minimum would occur when the two waves would differ by 1/2-wavelength in their path distance. This is equivalent to 0.305m. I drew a diagram and made a triangle in which one side was half of the distance between the speakers (2.925m), and the base was 0.305m. The angle adjacent to the 2.925m height and the unknown hypotenuse can be found by tan(angle) = 0.305/2.925, which gives an angle of 5.95 degrees. Since we wanted to know what angle is made with regards to the perpendicular bisector, we subtract 5.95 from 90 to find theta = 84.05.
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I hope my reasoning on this is correct. Thank you in advance.
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Close, but your triangle should have twice the dimensions.
If you draw the rays from both sources, both making angle theta with the perpendicular bisector, then the hypotenuse of your triangle would be the line between the sources. The longer leg of the triangle would run from one source to the ray produced by the other source. This ray would make angle theta with the hypotenuse.
The distance from this leg to the detector will be the same for both rays, since this leg makes a right angle (or very nearly so) with both rays.
The path difference would then be the third leg. Its length would be 5.85 meters (the hypotenuse of the triangle) multipled by the sine of theta.
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So 5.85 m * sin(theta) would be equal to half of the .61 meter wavelength, or .305 meter.
It follows that sin(theta) = .05, roughly, which corresponds to theta close to .05 (meaning .05 radians), or about 3 degrees.
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The perpendicular bisector of the segment connecting the two sources goes 'straight out' from that segment, and the two rays make angle approximately 3 degrees with the 'straight out' direction.
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