Your work on cylindrical lens 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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Distance from the back of the cylinder at which the band becomes thinnest, and diameter or circumference of the cylinder.
4.25, 12.70
The band was the thinnest at 4.25cm from the cylinder and the diameter of bottle was 12.75cm.
Distances at which the band focused most clearly and distances at which the focus lessened slightly, in order from closest to furthest; widths of the bands at these distances; distance of source from 'front' of cylinder; description of the bands.
6.4,7.4, 8.6
0.9,0.4, 1.1
109.4
At the closest distance (6.4cm) the line is the same shape as the clearer one (7.4cm). However, there are several differences in the bands. At the closest distance then middle portion of the band is not as bright as the outline. It has a fairly bright outline but the inner portion is a lot less bright. The band also looks less clear and more blurry. It is also bigger/wider.
At the middle distance (7.4cm) the line is the clearest. The line down the center is very distinct. At the top of the line there are two lines that form what looks like a fuzzy x A quarter of the way down the line there is a small blurry spot in the line. About half way down there there appears to be a fuzzy star shape. A quarter of the way to the bottom line there is a blurry spot with a couple of breaks in the line. The bottom portion forms a good clear line. It is by far the clearest and most focused of the three.
The farther distance is the fuzziest or most blurry. It was difficult to measure the width of the band due to the unclear lines. It doen't have a distinct shape like the closest and middle distances.
Width of the band; does the width of the band change linearly with position?
3.6cm
As you move the screen from the back of the cylinder to the point where the band is the thinnest and most in focus it gradually decreases in width. It appears to be linear. It is a very smooth transition. The further I moved the screen back the smaller the band became. At the back of the cylinder the width of the band was measured to be 3.6cm.
Past the point of narrowest focus, what happens to the width of the beam? How would you describe the region of space occupied by the beam?
Once you move back past the narrowest point of focus the bright region starts to widen again. However, when it widens the lines are difficult to see. They become extremely blurry and fuzzy. The edges of the band are no longer distinct and clear.
I would describe the region of space occupied by the beam as blank as a triangle leading the the index of refraction. It's wider at the cylinder and pointed at the point of clearest focus.
Index of refraction of water according to your measurements; details of your analysis and your analysis of uncertainties:
1.30
Using the formula (2-n/2n-2)r=actual distance I calculated the index of refraction. (2-n/2n-2)12.7=7.4
I calculated percent error (1.30-1.33)/1.33 *100=2.2% This uncertainty tells me I am 2.2 percent off what I should have observed. Due to the experimental errors I made I differed from the accepted value by 2.2 percent. 2.2 percent of 1.30 is .0286.
Index of refraction of second liquid.
1.43+- .03146
2.5, .15 width
2.2, .25 width
3.0, .25 width
Index of refraction=1.43
vegatble oil
distance was 101.5cm
The index of refraction is a little larger than water 1.43 as compared to 1.30 of water. The oil is thiker than water thus it refracts it more. Although, they are very close.
The second, third, and fourth lines indicate the clearest point, closest point of difference and furthest point of difference. I solved the index of refraction using the formula (2-n/2n-2)r=actual distance with my actual distance being 2.5cm. The radius of my cylinder was 3.75cm.
Index of refraction of a stack of CDs.
1.52 +- .0335
2.65, .25
2.35, .5
3.1, 0.6
The above three lines indicate the clearest point, closest point of difference and furthest point of difference. I solved the index of refraction using the formula (2-n/2n-2)r=actual distance with my actual distance being 2.65cm. The radius of my cylinder was 5.75cm.
How could you use the information in the first part of the experiment, where you measured the triangle, to determine the index of refraction of the light?
The triangular shape is the light rays converging on on the single point. It is the point where all of the light focuses together. We could calculate the index of refraction by drawing a line from the widest point of the triangle touching the cylinder then draw a line from that point to the center of the cylinder. Then draw a line from the center to the tip of the triangle. We could then solve for the anle of those lines by measuring the distances of the opposite over adjacent and taking the inverse tangent of that number.
We could also use the point where the light focuses the same way I did previously (2-n/2n-2)r=actual distance. The actual distance would be the distance from the point where the triangle converges to the cylinder. We just solve for n to find the index of refraction.
Index of refraction using halfway-screened cylinder, comparison between halfway-screened and unscreened cylinder.
1.30+-.029
water
The light best focuses at 7.3cm the band width there is 1.5cm
The closer measure ment that differe was 5.85cm and had a width of 1.75cm.
The furthest measure ment was 9.5cm away with a width of 1.8cm. The band looks very similar to the band before. It appeared the same shape but a little less bright for some reason. Using the formlua (2-n/2n-2)r=actual distance with an actual distance of 7.3 I found the index of refraction to be 1.30 +-.029.
How long did it take you to complete this experiment?
4 hours and 15 minutes
Optional additional comments and/or questions:
Excellent work, and excellent results.
My program inserts what it believes to be correct access codes when these are missing, based on the other info you give. The program isn't always right, but this time it was, so I just included my comments with the original submission.