cylindrical lens

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.

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.

7.0 cm, 10 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.

4.5 cm, 5.5 cm, 7.0 cm

5 mm, 2.5 mm, 6.0 mm

2.786 m, 2.796 m, 2.811 m

Width of the band; does the width of the band change linearly with position?

2.5 cm, The band appears to maintain the same brightness from center to edges.

The width of the bright region becomes thinner as it is moved to focus.

It appears that the width of the region changes lineraly with the position of the screen.

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 width of the bright region begins to increase as the screen moves past focus.

The region of space occupied by the beam after it leaves the cylinder is occupied by air.

Index of refraction of water according to your measurements; details of your analysis and your analysis of uncertainties:

1.33 +- 0.03

You don't show the details of your analysis or your uncertainties.

In Class Notes #17, so which the instructions referred you, the formula relating the position of focus, the radius and the index of refraction is developed.
The index of refraction n is related to the radius r of the cylinder and the distance d behind the cylinder at which the light focuses by
d = r·(2 - n)/(2·n - 2)
The question here is to find the index of refraction, having determined the distance d at which light focuses behind the cylinder, and the radius r of the cylinder. So solve the above equation for n we first multiply both sides by the denominator to get
d (2·n - 2) = r·(2 - n), then we expand to get 2 n d - 2 d = 2 r - r n. We rearrange to place all terms with n on the left, obtaining 2 n d + r n = 2 r + 2 d. We factor n out of the left side to get n (2 d + r) = 2 r + 2 d. Dividing both sides by 2 d + r we obtain n = (2 r + 2 d) / (2 d + r). Factoring 2 out of the numerator we have n = 2·(d + r)/(2·d + r).
If you plug your observed values of d and r into this equation you will get an equation for n.

Based on the 10 cm radius and the 7.0 cm distance at which the band becomes thinnest this formula yields
n = 2 * ( 7.0 cm + 10 cm) / ( 2 * 7.0 cm + 10 cm) = 34 cm / 24 cm = 1.42.
The accepted value of the index of refraction of water is 1.33, so your measurement is off by .08, or a little over 5%.
Your source was nearly 30 cylinder radii away; this could cause a change of a percent or so in the index of refraction, but this would also cause the distance of focus to increase, and would actually increase your error.

There was a little uncertainty in the measurements but it seemed to be minimal.

Index of refraction of second liquid.

1.40 +- 0.03

Vegetable oil.

What were the data on the basis of which you obtained this result?

Index of refraction of a stack of CDs.

Not sure what is being asked here.

The stack of CDs forms a short cylinder. The CDs, when viewed from the side, are pretty much transparent. Light will focus through this cylinder just as through any other cylinder.
From a distant source the focal distance will be on the same order as that observed for a large soft-drink cylinder, and is observed in the same way.

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 triagular shape was formed by the cylindrical shape of the cylinder with the apex being the light which passes through the center and the edges of the triagle formed by the outside of the cylinder.

The space would be air.

Not sure how I would determine the index of refraction.

The apex of the triangle should lie at a distance behind the cylinder which is very close to the focal distance.

Index of refraction using halfway-screened cylinder, comparison between halfway-screened and unscreened cylinder.

All the results from this portion of the experiment seemed to be the same as before. If there was any change it was not apparent to me.

The observations here should in fact not be significantly different than before.

How long did it take you to complete this experiment?

1.5 hours