Group3_8_ch24

Eric Solomon, Ross Dember, and Richie Johnson =Lab: Lenses= __Part A__ Focal Length 1/f=1/do+1/di 1/f=1/30+1/68.78 f=20.89cm

Percent Difference x and y intercepts (focal lengths) .0248*100 2.480%
 * x int - y int|/((x int + y int)/2)*100
 * 20.667-21.186|/((20.667+21.186)/2)*100

Avg. of x and y intercepts (focal lengths) (x int + y int)/2 (20.667+21.186)/2 20.927cm

__Part B__ 1/f using 1/do+1/di 1/f=1/do+1/di 1/f=1/50+1/36.2 1/f=.0476

1/f using 1/f(average) 1/f(avg) 1/20.9265 1/f=.0478

Magnification M=hi/ho M=-3/4 M=-.75

Percent Difference 1/f(avg) and 1/do+1/di .00340*100 .340%
 * (1/do+1/di)-(1/f(avg))|/(((1/do+1/di) + (1/f(avg)))/2)*100
 * .0476-.0478|/((.0476+.0478)/2)*100

Percent Difference M and -do/di .03528*100 3.528%
 * M-(-do/di)|/((M+(-do/di))/2)*100
 * -.75-(-.724)|/((-.75+(-.724))/2)*100

Part A Graph Values for graphs

x and y intercept for our graph Part B

The images are real when they are on the other side of the lens. In this case the image will be beyond the focus, where it will also be inverted. When virtual, it is on the same side of the lens, which would be upright and inside the focus. However, we were not able to see this because of the confines of the experiment. The ones that were tried behind the focus point were real and inverted. Those in front were upright, and virtual, yet unable to be seen in the experiment. There are two positions where it is focused because one will be beyond F and another will be between F and 2F. Since one is virtual you do not see or notice it. As shown in the data table for Part B, the image height was always negative (inverted image). As the image got closer to the lens, the image showed a trend of increasing in size up until we reached the object distance equaling the focal length. At the focal length, there was no image, and in front of the focal length, the image could not be viewed because it had become virtual and was therefore behind the lens. In simple terms, image characteristics of lenses and mirrors correspond inversely. This means that the image characteristics of a convex mirror are generally the same as those of a concave lens, and vice-versa. For a convex lens, like a concave mirror, the image can be either enlarged or reduced, virtual or real, and inverted or upright. This all depends on the distance that the object is from the lens/mirror. For a concave lens, like a convex mirror, the image is always virtual, upright, and reduced in size. For a lens, unlike a mirror, positive distance would be seeing the image through the lens. This would be negative when referring to a mirror**.** The opposite is also true. Therefore, real images would be through the lens, not reflected back off of them, as would be the case in mirrors. Trial four placed the object at the focal point of the lens. This produced no image. The light rays, after refracted by the lens, will become parallel to each other. In order to form an image, these rays must converge, and since they fail to do so, there is no image at the focal length. This is the same as in a concave mirror. For trial five, the image was placed between the focal point and the lens. The image did not and would not show on the viewing screen through the lens. This is because the produced image was virtual. Ergo, it was on the same side as the lens because the light rays did not converge through the mirror, but rather, on the same side as the mirror. Had we been able to view this image, it would have been upright, enlarged, and virtual. 1/di=1/f-1/do (in cm) 1/di=.05-.1 di=20 cm
 * 1. How do the slopes of the various lenses compare? How can you tell which lens has the bigger focal length using the graph?**
 * All the slopes, more or less, are equal to -1. You know what the bigger focal lengths are by the x and y intercepts, when the inverse of either the object distance or image distance is equal to zero, you know that the other one equals the inverse of the inverse focal length. **
 * 2. Are the images real or virtual? How do you know?**
 * 3. In all trials, were the images formed by the lens erect or inverted? Explain why.**
 * 4. Explain why, for a given screen–object distance, there are two positions where the image is in focus.**
 * 5. How did the image distance change as the object was brought closer to the lens?**
 * As object distance was brought closer, the image distance would get further, and vice-versa. **
 * 6. How did the image height change as the object was brought closer to the lens?**
 * 7. Compare and contrast characteristics and images of lenses and mirrors.**
 * 8. Explain your results for trials 4 and 5 in Part B.**
 * 9. For trial 5 in Part B, calculate the theoretical di, hi, and M. Show your work clearly.**

hi/ho=-di/do (in cm) hi=(-20/10)(4) hi=8cm

M=hi/ho M=8/4 M=2 No, image would be virtual and therefore not be able to be projected onto the screen because light rays would not intersect.
 * 10. Can you think of a way to project the image produced by an object between F and V onto a screen?**



=Lab: Refraction and Snell's Lab=

Top equation: water Second equation: acrylic Third equation: acrylic Fourth equation: water Fifth equation: water Bottom equation: acrylic Acrylic - This data shows the refracted angle measured for each incident angle, as light passed through the acrylic block, along with the sine measured for each measurement. We will use the sines to find the index of refraction for each value (reasoning in calculations). Water - This data shows the refracted angle measured for each incident angle, as light passed through the container of water, along with the sine measured for each measurement. We will use the sines to find the index of refraction for each value (reasoning in calculations). Class Data for Acrylic Class Data for Water - These two charts above show other group's results for the index of refraction and are used to help verify our results. Acrylic Template

Water Lens Template - The two pictures above show how we were able to measure the refracted angles, after marking where the light came out of the different materials.

[[image:snells_lab_2.0.png]]
2. Show why the slope of the line is the index of refraction of acrylic. Start with Snell’s Law, rearranging it to get it in the form of sin θ i vs sin ** θ ** r. The index of refraction for air is a known value, with n=1 (light moves at ever so slightly lesser of a speed than through a vacuum). Using the equation n1sin(theta1)=n2sin(theta2), the index of refraction of acrylic should be shown by this equation with n2 representing this value for acrylic: n2= (n1sin))/(sin(theta2)). Therefore, since the graph shows the value of sin(theta1)/sin(theta2), these values are the same. 3. If the materials are acrylic and water, what percent error do you get for each material? (If it is greater than 10%, you need to redo your data collection!) Acrylic- 5.50% (shown in calculations) Water- 2.38% (shown in calculations)
 * 1) Make a graph of sin θ i vs sin ** θ ** r. Include both lines on the same graph. Generate the equation of each line as well as the correlation coefficient. You may set the y-intercept equal to zero.

Conclusion Our hypothesis proved to be correct. After completing the experiment, the slope on our graph, representing, the index of refraction, came close to the class average and accepted value (using Snell's Law). Also, our values were all close to the trend line, showing that there was precision and accuracy in our results. Basically, this implicates a few things. First and foremost, it proves, along with our hypothesis, that Snell's Law is correct and applicable. Also, it shows the fundamental principle that the speed of light will change depending upon which medium it is traveling through. We presented this in our hypothesis through what is known as the index of refraction. It, in Layman's terms, tells how fast light travels through one substance compared to a vacuum. When we state that the index of refraction for water is 1.33, this means that light travels 1.33 times faster in a vacuum as it does in water.

Because of the simplicity of the experiment, the error is mostly human-based. For example, if the light beam was not perfectly on the line, the results would be skewed. Also, it would be possible for the block or tray to accidentally move during the experiment, which would throw off the results. However, there was little error in our results (5.50% and 2.38%), and that could have come from simply rounding off our measurements. Any other sources of error would have been, although present, extremely minor. We, for example, assumed that the water we used was purely water and contained no other substances. It probably had an ever so slightly different index of refraction from what we assumed. The same principle applies with the acrylic.

To fix the lab, the light beam could be placed so it was fixed at the different angles, so it would always be in the right position to measure the angles. Snell's Law has many useful purposes, such as for fiber optics, which are found in homes and responsible for devices such as television, as light is able to refract until it eventually reaches the destination, where it gives the light without any heat energy. Another point here would be to analyze the angles of refraction (compared to those of incidence) in relation to the normal line. This can show how the theories of optics apply in legitimate scenarios (for example: spear fishing). We can see for both water and acrylic, because they both require light to travel slower through them, that the mnemonic device "FST" applies. This means that since light travels from a fast medium to a slow medium, the angle of refraction will be closer to the normal line. This would mean that any time you are looking through different mediums, be wary that the image that you see is likely not in the place that it appears to be. =Lab: What is the relationship between intensity and angle of polarization?=

// As the angle of polarization increases towards 90º the intensity will decrease until there is not light at all. This relationship can be seen through the equation derived from Maul’s Law which is S=So*cos^2(ø). In this equation S is the intensity of light and So is the intensity of light after it passes through the first polarizing filter. Thus, as the angle increases the intensity decreases. Using this equation, we can determine that cos^2(ø) must be equal to the percentage passing through the neutral density filter. //
 * Hypothesis:**






 * Procedure:**

1. Set up two sources of light on either side of the light intensity measuring device 2. Set up a polarizing filter in front of one of the light sources 3. Set up a neutral density filter in front of the other 4. Turn off all other light sources so that no ambient light dilutes the experiment 5. Set a zero angle on the polarizing filter 6. Set the neutral density filter so that 100% of the light passes through it 7. Make sure that the intensities are of the same qualitative value 8. Systematically change the neutral density filter to each of the different percentages of light passing through 9. Match up the polarizing filter's angle so that the intensities match 10. Record the measured angles on the polarizing filter for each corresponding percentage value on the neutral density filter

The percent difference uses .75, .5, and .25 as the cos^2(theta) values respectively. This is because the transmittance percentage's corresponding decimal value must be equal to cos^2(theta) of the angle. This would give the correct percentage of light transmitted. Class Data Compared to the entirety of the class' data, we were quite close. In terms of percent difference from ours to the class data, we were (on the respective angles) 8.62%, 2.72%, and 1.67% off.
 * Data Table: **
 * % Transmittance || 75% || 50% || 25% ||
 * Trial 1 Angle (º) || 29 || 46 || 60 ||
 * Trial 2 Angle (º) || 29 || 46 || 60 ||
 * Trial 3 Angle (º) || 29 || 46 || 60 ||
 * Average Angle (º) || 29 || 46 || 60 ||
 * Cos2ø || .765 || .483 || .25 ||
 * % error || 1.99% || 3.49% || 0% ||


 * Discussion Questions: **

1. How do polarizing filters work? Polarizing filters take away a certain percentage of the light waves traveling through them. They are generally composed of two filters, one to take away an entire plane of light, and the other to take away a percentage of the perpendicular plane of light.

2. How is the polarizing filter different from the neutral density filter? Polarizing filters are more dynamic than neutral density filters. The neutral density filter blocks a set percentage of the light passing through it. The polarizing filter is a lined filter that can block a whole plain of light (which contains two plains) or any percentage in between 0 and 50% of light. It can be adjusted to different degrees so that a higher or lower percentage of the light is able to pass through.

3. How were your results? Our results were generally accurate. Assuming that (Cos(theta))^2 of the angle on the polarizing filter is equal to the percentage of light passing through, our results were on target.

4. Discuss sources of experimental error. We had very small error here. The obvious source would be that we were merely “eyeballing” the light intensity. We had no means of computing an exact value for the intensity comparisons. Therefore, our degrees would clearly be off by some percentage. The other main source was the fact that there was ambient light interfering with our results. The outside sources of light couldn’t be totally eliminated and some would pass through the filters, thus changing the results by a certain margin.

**Conclusion** Our hypothesis proved to be correct. As shown in our chart, each time the intensity decreased, the angle would get smaller. Because there was more light passing through the filter, there needed to be more light "cut-off" in order to keep the light the same. Our results were consistent the whole time, even when distances were changed, which showed how the relationship and intensity remained the same regardless of other factors that contribute to light.

As discussed in the questions above, our error was very small and the errors came from ways of measuring. For example, the was no quantitative way of measuring if the light strips were the same, and measuring the angle was not exact. Also, there could have been other light getting into the filter that could have affected our results. However, given our accuracy, the sources of error did not have too big of an influence.

Other than academic reasons, this lab had real-life applications. For example,, buildings have technology that allows them to allow some natural light in and keep some of it out in order to allow the building to use less electricity. This technology is located on windows, which have filters on them, so when there are clear skies, which means the intensity of light is high, the angle of the filters will not be as big as if the people in the building wanted the same light during the day. Polarizing filters are then ecologically beneficial and economically effective.