Your 'cylindrical lens' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
Your optional message or comment:
Distance from the back of the cylinder at which the band becomes thinnest, and diameter or circumference of the cylinder.
10.3, 4.8 cm
diameter
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.
3.8, 3.2. 2.11, 4.1
.15, .18, .25, .22
15.2 cm
The band just appeared to get larger and larger.
Width of the band; does the width of the band change linearly with position?
.42
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?
The brightest region becomes larger and larger. The actual space of the beam of light is much larger than the space it was coming from aka the source it was coming through.
Index of refraction of water according to your measurements; details of your analysis and your analysis of uncertainties:
The index of refraction is defined as the incident wave on the interface between two media with opposite refractive index signs completely being transmitted. There is no reflective wave and can only happen with the two materials have a negative refractive index.
1.36, 1.33, 1.35, 1.39
+ - .01
It's not completely clear what these numbers represent.
Your index of refraction for water, according to your measurements, would be based on the relationship
d = r * (2 - n) / (2n - 2), where r is the radius of the container and d the distance behind the container at which the beam best focuses. To get n solve the equation:
2 d(n-1) = r(2-n) so
2d n - 2d = 2 r - r n so
rn + 2 d n = 2 r + 2 d so
n ( r + 2 d) = 2(r + d) so
n = 2(r+d) / (r + 2d).
If r = 1/2 * 10.3 cm and d = 3.6 cm (fairly close to the average distance you observed for water) we get
n = 2 ( 10.3 / 2 + 3.6) / (10.3 / 2 + 2 * 3.6) = 1.42.
Index of refraction of second liquid.
5.8 +- .25
Clear vegetable oil
The data was a measurement of the refractiion where I was looking at the vegetable oil from an angle or slant.
If d = 0--i.e., the focus is directly behind the cylinder--we get n = 2. If the focus is anywhere beyond that--at any finite distance behind the cylinder--then n < 2. So n = 5.9 is not possible.
Can you clarify the data you obtained for the oil?
The same is the case for the CDs below.
Index of refraction of a stack of CDs.
6.9 +- .14
DVDs
Again for this part of the experiment, the slant was just slightly higher than previously measured. Here the refraction was larger because the angle was larger coming from a higher position.
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 triangle is almost bent. The sides come in at a bent curve nature. The region of space is somewhat small considering the amount of light produced. If I were to measure the amount of light produced, I think this would be a less active way because there can be ambiguous measurements made.
Index of refraction using halfway-screened cylinder, comparison between halfway-screened and unscreened cylinder.
2.25 +- .2
concentrated sugar water
This was slightly smaller than the vegetable oil because it was much like the cylinder with just water. This vegetable oil was thick and oily creating a more widened image. The image simply went through the water.
How long did it take you to complete this experiment?
Optional additional comments and/or questions:
The data you report are reasonable. Your values for the index of refraction are not all reasonable. This should be easy to clarify. See my notes and see if you can either modify your results or clarify your data.
Your 'cylindrical lens' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
Your optional message or comment:
Distance from the back of the cylinder at which the band becomes thinnest, and diameter or circumference of the cylinder.
10.3, 4.8 cm
diameter
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.
3.8, 3.2. 2.11, 4.1
.15, .18, .25, .22
15.2 cm
The band just appeared to get larger and larger.
Width of the band; does the width of the band change linearly with position?
.42
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?
The brightest region becomes larger and larger. The actual space of the beam of light is much larger than the space it was coming from aka the source it was coming through.
Index of refraction of water according to your measurements; details of your analysis and your analysis of uncertainties:
The index of refraction is defined as the incident wave on the interface between two media with opposite refractive index signs completely being transmitted. There is no reflective wave and can only happen with the two materials have a negative refractive index.
1.36, 1.33, 1.35, 1.39
+ - .01
It's not completely clear what these numbers represent.
Your index of refraction for water, according to your measurements, would be based on the relationship
d = r * (2 - n) / (2n - 2), where r is the radius of the container and d the distance behind the container at which the beam best focuses. To get n solve the equation:
2 d(n-1) = r(2-n) so
2d n - 2d = 2 r - r n so
rn + 2 d n = 2 r + 2 d so
n ( r + 2 d) = 2(r + d) so
n = 2(r+d) / (r + 2d).
If r = 1/2 * 10.3 cm and d = 3.6 cm (fairly close to the average distance you observed for water) we get
n = 2 ( 10.3 / 2 + 3.6) / (10.3 / 2 + 2 * 3.6) = 1.42.
Index of refraction of second liquid.
5.8 +- .25
Clear vegetable oil
The data was a measurement of the refractiion where I was looking at the vegetable oil from an angle or slant.
If d = 0--i.e., the focus is directly behind the cylinder--we get n = 2. If the focus is anywhere beyond that--at any finite distance behind the cylinder--then n < 2. So n = 5.9 is not possible.
Can you clarify the data you obtained for the oil?
The same is the case for the CDs below.
Index of refraction of a stack of CDs.
6.9 +- .14
DVDs
Again for this part of the experiment, the slant was just slightly higher than previously measured. Here the refraction was larger because the angle was larger coming from a higher position.
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 triangle is almost bent. The sides come in at a bent curve nature. The region of space is somewhat small considering the amount of light produced. If I were to measure the amount of light produced, I think this would be a less active way because there can be ambiguous measurements made.
Index of refraction using halfway-screened cylinder, comparison between halfway-screened and unscreened cylinder.
2.25 +- .2
concentrated sugar water
This was slightly smaller than the vegetable oil because it was much like the cylinder with just water. This vegetable oil was thick and oily creating a more widened image. The image simply went through the water.
How long did it take you to complete this experiment?
Optional additional comments and/or questions:
The data you report are reasonable. Your values for the index of refraction are not all reasonable. This should be easy to clarify. See my notes and see if you can either modify your results or clarify your data.