Group1_8_ch24

Lab Group 1 Erica Levine, Bret Pontillo, and Steven Thorwarth

toc =Lab: Polarization= Date: 12/19/11

__Purpose:__ To find a relationship between the intensity of light and the angle of polarization for a point.

__Hypothesis:__ We will find an indirect relationship between light intensity and the angle of polarization. We can determine this because of Malus' Law, which is represented by the equation S = S o * cos 2 ø where S is the intensity of light, S o is the intensity of light after the first filter, and ø is the angle between the axes of the two filters involved. From this equation we know that as the angle ø between the two filters increases, the intensity of light that shines through (S) decreases.

__Materials:__
 * 1 Optics Bench
 * 2 Point Light Sources
 * 1 Light Polarizer
 * 1 Neutral Density Filter
 * 1 Photometer

__Procedure:__
 * Place the optics bench on a flat surface, preferably in a dark room with no windows.
 * On the optics bench, place two point light sources at opposite ends with the point lights facing one another.
 * In the middle, place a photometer with the two ends facing the point light sources.
 * On one end of the photometer, directly next to the photometer, place a light polarizer, on the other end attach the neutral density filter to the photometer.
 * Make sure the neutral density filter is set to 100%.
 * Turn on the point light sources and turn off the lights in the room.
 * Next, adjust and record the angles of the light polarizer so that the photometer indicates that the light intensity from both ends are equal.
 * Then, switch the neutral density filter from 100% to 75%, and adjust the angle of one end of the light polarizer until the photometer reads equal intensity for both ends.
 * Measure the angle required to do this on the polarizer and record the change in angle.
 * Repeat the previous two steps to measure when the neutral density filter is set to 50% and 25%.
 * In order to assure precise data, repeat this procedure 3 times, getting three points of data for each % on the neutral density filter.

__Data:__ OUR RESULTS



CLASS RESULTS

__Calculations:__

__Discussion Questions:__

1. How do polarizing filters work? Polarizing filters block one of the two planes of vibration of an electromagnetic wave. The filter will take out a percentage of the vibrations depending on its angle, which dims the light you see.

2. How is the polarizing filter different from the neutral density filter? A neutral density filter is merely a dark piece of glass that reduces the amount of transmitted light. On the contrary, a polarizing filter has lines of narrow grating, which only allow 50% of light to be transmitted. If a second polarizing filter is used, it can block the remaining 50% of light depending on the angle it is placed at in relation to the first polarizing filter.

3. How were your results? Our results were fairly accurate, as seen above by our percent difference. Our highest percent difference from the class averages was 9%, and our results were relatively close to those of the class. Our results also confirmed our hypothesis that as the angle between the two filters increased, the amount of light transmitted decreased.

4. Discuss sources of experimental error. Human error is huge part of the error within this lab. It was difficult to read the angles projected on the filter and also trying to match the exact brightness on both sides of the optics bench. Error could have also come from other light in the classroom. We tried to account for this on one of the light sources which projected out of both sides and to the ceiling. We covered the one going to ceiling in hopes to obtain better data. Although, the room was not completely dark.

__Conclusion:__

Based on our results, we were able to conclude that our hypothesis was correct. As the angle between the two polarizing filters increased, the amount of light transmitted decreased. The amount of transmitted light is directly proportional to the cosine squared of the angle. However, as the angle increases, the cosine squared decreases, making the relationship between transmitted light and angle an inverse one. The answer comes out as a decimal, which can then be multiplied by 100 to find the percent of light transmitted. Our results were fairly accurate, based on the percent difference from the class averages. Our percent differences were fairly low, with out largest being 9%. Sources of error were already addressed above, but in the future it would definitely be necessary to make changes to the design of the experiment in order to have more accurate results. In the future it would be beneficial to find a mechanical way to determine when the brightness of the two lights were equal, because human error could have contributed to the imperfection of our results. In addition, it is important to find a more precise method of measuring the angle between the two polarizers, and also to make sure that no additional light was coming from anywhere in the classroom. There are many applications of this lab in real life. The idea of polarized lenses are often used in photography and the creation of sunglasses. In both areas, polarized lenses are used to block a certain percentage of transmitted light.

=Lab: Snell's Law= 1/5/12

__Purpose:__To derive Snell's Law experimentally, to determine the index of refraction experimentally, and to compare the index of refraction for acrylic and water.

__Hypothesis:__ If we graph the sin of the angle of refraction vs. the sin of the angle of incidence for a certain medium, the slope of that line will be the index of refraction for the respective medium. We formulated this hypothesis based on the Snell's Law Equation which states that. We know that the index of refraction for air is 1, so the the n2 will be the ratio (or slope) of the angle of refraction over the angle of incidence.

__Procedure & Materials:__ First we placed the rectangular acrylic block on the lab table on the paper given to us. We then shined the light source through it at various angles. For each angle, we clearly marked the position where the light EXITS the block. Then we measure the angle of refraction each time with a protractor. Then repeat the procedure using water lens. We then used a ruler to connect draw the lines INSIDE of the box, and a protractor to measure each angle or refraction.

Acrylic-



Water-

__Data:__ __Analysis:__

This graph has two lines graphed. One is the Sin (angle of refraction) vs. Sin (Angle of Incidence) of acrylic and the other is of water. As shown in the calculation below this, the slope of the equation gained from our graph is index of refraction for that medium. For acrylic, there is a slope of 1.468. This had an accurate R^2 value also of .991, showing the accuracy of our line. For water, there is a slope of 1.351, with an R^2 value of .998 which is also a highly accurate value.

slope of sinø i v. sinø r:



Percent Error and Percent Difference:



__Discussion Questions:__ 1. State as quantitatively as possible the precision of your value for //n//, the index of refraction.
 * Our values for n are extremely precise, with measured values of 1.351 and 1.468, we had a percent error for both acrylic and water of about 1.5%. For water we got an n value of 1.351 which, when compared to the actual value of 1.33, has a percent error of 1.579%. For acrylic, we measured an n value of 1.468 which, when compared to the actual value of 1.49, has a percent error of 1.477%. These are very low percentages, indicating that our results were extremely accurate and precise.

 2. State how your data for the acrylic prism are evidence for the validity of Snell’s law.
 * Through this experiment we are able to test Snell's law with an acrylic prism. We shined a light through the acrylic cube at various angles and measured their corresponding reflected angle for the lights path. Then, by plotting our values of sin(angle of incidence) v. sin(angle of refraction) we find a linear equation for our data where the slope is the index of refraction of the acrylic. We know that the slope is the index of refraction based on Snell's Law, as we rewrite it as sinøi = (nr/ni)sinør, where ni is the index of refraction for air (1), and so it equals sinøi = (nr)sinør. The slope of our graph, our measured index of refraction, was 1.468, which is very close to the actual index of refraction of 1.49. Because our value is so close to the actual value for the index of refraction for acrylic, we prove Snell's Law to be true and valid.

 3. Using the value of //n// determined for the acrylic block, find the speed of light in the prism.

 4. Inside a prism the wavelength of the light must change as well as the speed. Is a given wavelength longer or shorter inside the prism? Consider specifically light whose wavelength is 500 nm in air. What is the wavelength of this light inside the prism? 

__Conclusion:__ Our hypothesis, that the graph of the sin of the angle of refraction vs. the sin of the angle of incidence for a certain medium, would be the slope of that line will be the index of refraction for the respective medium was correctly proven by the results of our experiment. When we created the graph for an acrylic prism, the slope of the graph, 1.468. The known index of refraction of acrylic is 1.49, so our experimental value obtained in our graph was right on point with the actual. For water, we had an index of refraction of 1.351 which was the slope of the line in our graph. The actual index of refraction for water which is 1.33, so once again our experimental was right there with the actual index of refraction.

As for error, there was not much in this lab. Using the resulting slope from the acrylic portion of this lab (1.468), we were able to compare it to the actual (1.49) and find a percent error of only 1.477%. As for the water portion, we had a percent error of only 1.579% using the theoretical (1.351) and the actual (1.33). The error that we did get occurred in how we measured the angles of incidence and the angle of refraction. Because we could not directly and accurately measure these angles on the acrylic or water, we had to draw the points where the light entered the surface and where it left the other end. Then we connected the dots and measured the angles. To correct this error, we would need a better way to measure the angle of reflection along the surface of the tested object. This could be done by having a portion of the acrylic, or the water container, that directly measures angles on the surface. The other source of error is the measurement of the angle in which the light approaches the surface before being reflected. The light source was free standing and placed by the naked eye. This could cause inaccuracies in the angle of incidence and an error in our resulting angle of refraction. To fix this error, we should place the light source on a stand in which angles of incidence are measured directly, in order to assure that the light source is angled the way it is supposed to be.

This concept is important to know for allowing light through surfaces in the most efficient way possible. This may not seem important, but it is a major factor in natural lighting with sunlight through windows, which when done efficiently can save large amounts of electricity in lighting during the daytime. Another use of this object is within fiber optics, which is the now used within the Verizon television and phone line provider. This is important because within fiber optics the refracted angle does not leave, and can go long distances. Fiber optics are also important to understand because in comparison with copper wire which is typically used with television wiring outside homes, the fiber optics do not heat up, whereas the copper wire does. If the a wire would heat up it could bend more, which could make a home lose connection.

=Lab: Lenses= 1/12/12

Purpose: Using several different measured object lengths and their corresponding image lengths, we will find the focal length of a lens, as well as verify the characteristics of the image (location, height, virtual or real, inverted or right-side up) for each object distance. This data will be used to prove that the thin lens equation is true.

Hypothesis and Rationale: By plotting the inverse of the object distances and the inverse of the corresponding image distances on a graph, the inverse of the x-intercept and the slope of the resulting graph will be equal to the focal length of our lens. We know this based on the equation 1/f=1/di + 1/do.

Methods and Materials: To do this experiment, we began with an optical bench which we placed on a flat surface (our desk). Then, we attached our object (the light source) to the optical bench where distance=0cm. It is important that we aimed the light source so that the image (ours was two parallel arrows) down the optical bench. This allows us to measure the object distance with ease. Then we placed a blank white screen on the opposite end of the optical bench from the light source where the image was facing. Finally, we placed a thin lens holder in the middle, and turned on the light source by plugging it into a power outlet. In order to get the data necessary, we placed the lens holder in a position that is between the light source and the screen, and moved the screen to a position where there was a clear image that is similar to the object on the light source (the image should be VERY CLEAR). We measured and recorded the distance of the lens holder from the light source(D o ) as well as the lens holder from the screen(D i ). Then we moved the lens holder to an other position between the light source and screen, found where the new image appeared with the screen and recorded the same measurements. We repeated this process for 10 specific and different locations of the lens holder, using these points to generate our graph.

Data: Part A: Group Pontillo, Levine, Thorwarth Lens- 250 mm Group Listro, Rabin, Irwin Lens- 100 mm Group Fihma, Hallowell, Litmanov Lens- 500 mm Group Johnson, Dember, Solomon Lens- 200 mm

Part B data:

Graph: Calculations: Finding the focal length you using the graph. The following calculation is also used for the y intercept.

F from y is the Focal length found using the y intercept and F from x is the focal length found using the x intercept.

Analysis: Discussion Questions:
 * 1) How do the slopes of the various lenses compare? How can you tell which lens has the bigger focal length using the graph?
 * 2) the slopes of the lenses were all around a similar value of -1. The way that you can tell what the focal length is is by looking at the x- intercept of each graph. The x-intercept is the inverse of the focal length for that lens.
 * 3) Are the images real or virtual? How do you know?
 * 4) The images are real when the object is located behind the focal point. We can tell they are real based on the location of the image relative to the lens and the object (light source). Because the image is on the opposite side of the lens from the object, we know it is real. The images are virtual when they are located in front of the focal point. We can tell they are virtual based on the location of the image relative to the lens and the object (light source). The images are virtual when they are located on the same side of the lens as the object.
 * 5) In all trials, were the images formed by the lens erect or inverted? Explain why.
 * 6) The images that were real appeared to be inverted. These were the images formed when the object was beyond 2F, at 2F, and between 2F and F.The image that was virtual appeared to be upright. This image was formed when the object was located in between F and V. This can be validated using the equation hi/ho=-di/do. When object and image distance are both positive, then the image height will be negative (or inverted). When the image distance is negative, the resulting image height will be positive (upright).
 * 7) Explain why, for a given screen–object distance, there are two positions where the image is in focus.
 * 8) This is because, for a convex lens, there are two focal points on each side of the lens (F and 2F). We know that an image for a specific screen-object distance will be created at some object-lens distance between F and 2F, as well as beyond 2F. We can eliminate distances before F because they create an image behind the object, and at F creates no image.
 * 9) How did the image distance change as the object was brought closer to the lens?
 * 10) As the object distance gets smaller, the image distance gets larger.
 * 11) How did the image height change as the object was brought closer to the lens?
 * 12) As the object distance gets smaller, the image size gets larger.
 * 13) Compare and contrast characteristics and images of lenses and mirrors.
 * 14) Images in concave mirrors follow the same principals as images in convex lenses. Images in convex mirrors follow the same principal as images in concave lenses. Images in concave mirrors and concave lenses are both virtual, upright, and reduced in size. Images in convex mirrors and concave lenses depend on the location of the object. When the object is behind the focal point, the images are real and inverted. When the object is at the focal point, no image is seen. When the object is in front of the focal point, the image is virtual and upright. The magnification of images in convex mirrors and concave lenses depend on the location of the object.
 * 15) Explain your results for trials 4 and 5 in Part B.
 * 16) In trial 4, no image was found because the object was located at the focal point. In trial 5, the image was located behind the lens because the image is virtual when the object is between the vertex and the focal point.
 * 17) For trial 5 in Part B, calculate the theoretical di, hi, and M. Show your work clearly.
 * 18) [[image:Screen_shot_2012-01-15_at_3.18.36_PM.png]]
 * 19) Can you think of a way to project the image produced by an object between F and V onto a screen?
 * 20) A lens with a half screen could be used to produce an object between F and V onto a screen. The image is located on the same side of the lens as the object, and the light shines through the open half of the lens, the image will be projected onto the covered half of the lens.

Conclusion:

Our data proved our hypothesis from Part A to be correct. We graphed the inverse of the object distance vs. the inverse of the image distance, and the resulting slope was the inverse of the focal length. We were able to find this relationship based on the thin lens equation, which states that 1/f=1/do + 1/di. Our experimental focal length was 25.52 cm, while the theoretical was 25 cm. The two values were very close, with a percent error of only 2.08%. A possible source of error in this part of the experiment was the reading of the image. It is possible that the the image distance that we recorded was not the exact distance at which the image was most focused. However, this only accounted for a minimal amount of error. In the future, it would be better if we had a device that could read where the image was most focused.

Our data also proved our hypothesis from Part B to be correct. Images in convex lenses follow the same pattern as images in concave mirrors. When the object was beyond the 2F point, the image was real, inverted, and smaller. When the object was at the 2F point, the image was real, inverted, and the same size. When the object was between the focal point and the 2F point, the image was real, inverted, and larger. When the object was located at the focal point, no image existed because there is no point where the rays intersect. Lastly, when the object was in front of the focal point, we were not able to see it. This is because the image was virtual. This image would have been located behind the lens, would have been upright, and would have been larger than the object. We can use calculations to find the exact image height and image distance of this virtual image. We calculated the magnification of the image based on heights and distances. The percent difference between these two values was only 7.94 percent, showing that our results were fairly accurate. Once again, our error most likely came from human error and the minimal error from Part A. Lastly, we found another way to validate the thin lens equation. We projected the image of the window onto the screen from a very far distance, and we measured the object distance to be 25 cm. If the thin lens equation is 1/f= 1/do +1/di, and do is approaching infinity, then 1/do is approaching zero and 1/f must equal 1/do. We know the thin lens equation was correct because di came out to be 25 cm, the exact value of our focal length.

A real life application of this lab is contact lenses and glasses. Optometrists use this the knowledge we explored in this lab every day to help create personalized lenses and glasses that will improve the sight of their patients.