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course Phy 122
3/31 10:23 pm
Experiment 29: InterferenceUsing a hand-held laser pointer and a diffraction grating consisting of lines on a rectangular transparency, we observe the maxima created when the light is directed through the pattern at various separations, and with various incident angles. We determine the angular separation of the maxima and use this separation to estimate the wavelength of the light. We then use sets of parallel straight lines on the same transparency to determine the wavelength of the light.
Stapled to the paper rulers in your lab materials package is a rectangular transparency a few inches on a side. The transparency contains copies of various patterns of lines.
In at least one pattern the lines form a V.
Orient the pattern so that the V is upright, with the widest spacing at the top.
Move at least 3 and preferably 5 or more meters from a smooth wall. Shine the laser through the V near the top of the pattern and observe the image made by the light on the wall. Measure the distance from the transparency to the wall.
Gradually move the laser down through the V, so that it shines between lines that move progressively closer and closer together. Observe what happens to the pattern on the wall.
Continue moving down the V until you obtain the most distinct possible set of bright spots on the wall.
Note the vertical position of the beam on the V.
As best you can, determine for this position the average distance between the distinct bright spots formed on the wall.
Measure the width of the V at this point, and the number of spaces between the threads across the width.
Record also the distance to the wall.
I was unable to get any kind of reading with this method. The laser was too blurry on the wall and any kind of set up that allowed gradual movement, did not allow for any kind of measurement.
Perhaps this was because my laser pointer was not particularly good quality. I've had it for a while, it was a token gift I'd gotten at a conference, and I only used it for giving PowerPoint presentations where precision wasn't a real necessity.
I had some limited success with the rectangular patterns and those results are included below.
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There are also a few rectangular patterns consisting of parallel lines. The spacing of the lines varies from rectangle to rectangle.
Repeat the preceding exercise using different rectangular grids.
For each grid determine the average distance between the bright spots on the wall, the average distance between the grid lines and the distance from the plastic rectangle to the wall.
For this part, I used a ladder to create a stable platform for the laser and then with some tape and some tea lights as holders, I kept the transparency rather stable. The dots on the wall were never distinct dots. It was basically a blob with black lines through it. Even so the lines were not clearly defined and often the lines were not uniformly distant so I would not have much confidence in my measurements.
Distance was kept at 455 centimeters which was the best spot to position the ladder between 3 and 5 meters.
For the grid spaced at .5mm the lines in the blob were spaced at 2mm
For the grid spaced at 1mm the lines in the blob were spaced at 3mm
For the grid spaced at 2mm the lines in the blob were spaced at 3mm
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According to your results, how is the spacing between the bright spots on the wall related to the distance between the lines?
The results are inconclusive.
Just using the .5mm and .1 mm lines I would say that if you double the line spacing on the rectangle, you increase the line spacing on the wall by a factor of 1.5. We'd have to throw out the 2mm spacing results though.
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What is the ratio of the spacing between the dots to the distance between the plastic rectangle and the wall?
For the .5mm lines the ration of spacing to distance is .2cm/455cm or .000440
For the 1 mm lines the ratio of spacing to distance is .3cm/455cm or .000659
For the 2 mm lines the ratio of spacing to distance is .3cm/455cm or .000659
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What distance is in the same ratio with the spacing between the lines?
???I don't understand this question???
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For example:
If the spacing between the lines is .5 mm and the ratio is ..000440, then the distance in the same ratio with the spacing is .000440 * .5 mm = .000220 mm, or 2.20 * 10^-7 m.
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&#This looks good. See my notes. Let me know if you have any questions. &#