Group2_8_ch24

Sam Fihma, Chris Hallowell, and Phil Litmanov
flat =Polarization Lab=

To find the relationship between intensity and angle of polarization.
 * Purpose:**

As the angle of polarization increases from 0 towards 90, the magnitude of the resulting intensity will decrease. This means that they are inversely proportional from 0 to 90 degrees. As the angle increases from 90 towards 180, the magnitude of the resulting intensity will increase as well. This means that they are directly proportional from 90 to 180 degrees. We developed this hypothesis based on the equation,. I was able to use some example angles to see that from 0 to 90 degrees, the angle of polarization is inversely proportional to intensity. I was also able to find out that when the angle increases from 90 towards 180, the angle of polarization is directly proportional to intensity.
 * Hypothesis:**

The materials we used in this lab were an optics bench, a point light source, 2 polarizers, a photometer, an aperture, and 3 brackets.
 * Materials:**


 * Diagram of Set-Up:**

Our Data
 * Data**:
 * The "% Error" is comparing our "(cos(theta)^2)*100 " value to the original "% transmittance" value.
 * The "% Difference" is comparing our average angle to the class average angle.
 * The data table is in 3 significant figures.

Class Data


 * Calculations:**









We redid the experiment a few times just to improve our results. Our findings are much better than they were when we first did the trial. For 75% transmittance our error was 0.67%, 50% was 8.14%, and 25% was 21.3%. The sources of error are explained below. We were pretty close to what the class average in each percent transmittance was. We had 3.77%, 5.61%, and 7.61% difference, respectively.
 * Analysis:**


 * Discussion Questions:**
 * 1) How do polarizing filters work?

A polarizing filter works by blocking out one of the two planes that light vibrates on. In effect, a polarizing filter blocks out 50% of light going through it.


 * 1) How is the polarizing filter different from the neutral density filter?

A polarizing filter always blocks one plane of vibrations so it always blocks 50% of the light. A neutral density filter can block any percent of light from 1-100%. A neutral density filter has a blocks light by being composed of black spots. A 25% neutral density filter allows 25% of light through and is composed of 75% of black spots.


 * 1) How were your results?

Our results were quite accurate. We had relatively small error everywhere. Our highest error was for 25% transmission.


 * 1) Discuss sources of experimental error.

Sources of error from this lab result from our own inaccuracies. It was almost impossible to know exactly what angle was being read, and so we had to take an educated guess. We also had to guess whether or not the two lines were the same brightness. Sometimes certain group members disagreed on the brightness of the lines and so we had to guess when they were equal. Lastly, because the room wasn't completely dark, the light could have thrown off our observations and data.

Our hypothesis was somewhat correct in that it did identify how the intensity reacts when the angle is changed. As the angle gets closer to 90 degrees, the intensity of the light gets weaker. As the angle gets closer to 0 degrees, the intensity of the light gets stronger. Thus, as the angle becomes bigger, intensity weakens and vice versa. This would known as an inverse relationship normally, but it is a bit more complicated than that. The cosine of the angle needs to be taken and then it needs to be squared (this equation is the same one used in our hypothesis). The bigger the angle, the smaller this number will be and vice versa. The number that we get from doing this has a direct relationship with intensity. This number represents the percentage of light that is transmitted (it can never be over 1). For example, if this number is .75 and it is multiplied by the original intensity of the light source, the new intensity will just be 75% of the original. Error was just explained above in the discussion questions, but there are definitely ways to address it. A better way to measure the angle would be helpful in this lab. I felt that our angles were not very accurate multiple times through the lab. We would also need a better device to make the two intensities match up. The color scheme idea did not work too well. This is relevant in real life through everyday objects, such as windows and sun glasses. Physicians who engineer pieces of glass need to have a certain percent transmission for what they are working on and they use things we did in our lab to figure out how to build objects.
 * Conclusion**

=Lab: Refraction and Snell's Law=

The purpose of this lab is to experimentally derive Snell's Law, experimentally determine the index of refraction, and to compare the index of refraction for acrylic and for water.
 * Purpose:**

When graphing the sine of the angle of incidence vs. the sine of the angle of refraction, the slope of the line will be the index of refraction for acrylic/water. Our rationale is Snell's Law:
 * Hypothesis:**



n1 is the index of refraction for air which is 1.0 so:



Following the equation for a line, y = mx + b, it is clear that y = sin(theta1) and x = sin(theta2) and b = 0, and m (slope) = n2.

n2 = index of refraction of acrylic/water

The materials used in this lab were an optics bench, light source, acrylic block, water lens, protractor, and ruler.
 * Materials:**

We began by putting the acrylic prism on the template that was provided for us by our teacher. Next, we shined a single beam of light through the prism at various angles. Then, we marked the point where the beam of light escaped from the other side of the prism and drew a line with a ruler from where the light entered, to where the light exited. We then used a protractor to determine the angle of refraction for each of the lines. Finally, we recorded each of the angle values in our data table. We performed this same procedure using a box filled with water as well.
 * Procedure:**

Acrylic
 * Photos of Templates Used:**

Water

The following data is for the coefficients of each group's data.
 * Data:**


 * Graph:**


 * Analysis:**
 * We showed in our hypothesis how the slopes of the lines are the indices of refraction for acrylic and water.

Percent Error for Acrylic
 * Calculations:**

Percent Error for Water

Percent Difference for Acrylic (Compared to Class Average)

Percent Difference for Water (Compared to Class Average)


 * Discussion Questions:**

1. State as quantitatively as possible the precision of your value for //n//, the index of refraction.

Our values for //n// are very precise. Both of our percent errors were about 2%. Our experimental value for acrylic is 1.5202 and the theoretical value is 1.49, our percent error for water is 2.03%. Our experimental value for water is 1.3672 and the theoretical value is 1.33, and our percent error for acrylic is 2.8%. Our experimental results are very near the theoretical values, which means our values for //n// are very precise.

2. State how your data for the acrylic prism are evidence for the validity of Snell’s law.

Our data for the acrylic prism is evidence for the validity of Snell's Law due to the fact that our percent error was so low. As seen above, the percent error turned out to be only about 2.03%, which shows that our experiment proved, almost exactly, that Snell's law is, in fact, true. In addition, the derivation, which we showed in the rationale for our hypothesis, explained why the information our graph presented helps us to prove why Snell's law is 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?

The given wavelength is shorter inside the prism.


 * Conclusion:**

After performing the lab, we can conclude that our hypothesis was, in fact, correct. Our data along with our minimal percent error values show that we were able to validate Snell's law effectively. After showing how the Snell's law equation could be seen in our graph, it was evident that we had collected today that would prove our hypothesis correct. The slope of the line for the acrylic graph showed that our experimental value for the index of refraction was 1.5202, while the actual value is 1.49. This shows that we almost validated the law perfectly. The slope of the line for the water graph showed that our experimental value for the index of refraction was 1.3672, while the actual value is 1.33. This, once again, showed our hypothesis was correct.

As seen in our calculations above, the percent error for the acrylic part of the lab was 2.027%, while the percent error for the water part of the lab was 2.797%. This shows that we did a pretty good job of collecting our data and performing the lab, however, there were still a few steps during our process where we could have made a slight mistake. First, when we were marking where the beam of light was exiting the acrylic/water, it was tough to tell exactly where to mark. This was because when the beam of light exited, it spread out and was not a fine, focused beam, which made it tougher to get the perfect mark. Another possible source of error could have came when we were measuring the angles of refraction using the protractor. Because the marks on the protractor are so small, it made determining the exact angle a bit tougher.

Overall, in the future, I would change the lab by using a stronger beam of light. This would help to get a more exact exit point from both the acrylic prism and the water. In addition, I would also try to use a device that would measure the angles of refraction with a bit more precision to ensure that we are obtaining the exact values. A real-life example of this concept and idea is seen anytime you look into a fishbowl. Due to the concept of refraction, if one looks into the fishbowl at an angle, the fish will seem to be in one place, while it is really in another place.

=**Lab: Lenses**=


 * Purpose:**

To precisely determine the focal length of a thin lens. To verify the image characteristics formed when objects are placed at varying distances from a lens. To validate the thin lens equation.


 * Hypothesis and Rationale:**

Part A is centered around one equation shown below. Our observations and analysis should support this equation. We know this to be true after reading lessons from the Physics Classroom and work done in class. Part B is all about the characteristics of images at certain points in comparison to the lens. When an image is created during this part, it will be real and inverted. However, there will be no image created at the focal point. There will not be an image seen in this experiment when the object is between the focal point and lens because it will be virtual. We know this because of our reading done on the Physics Classroom.

(All materials are bolded) Part A:
 * Methods and Materials:**

To start, attach the **light source** to the **optical bench**, with the arrow side toward to screen. Next, position the **convex** lens between the light source (the object) and the **screen**. When positioning these pieces, be sure the object and the screen are at least one meter apart. At this point, one must slide the lens to a position where an image of the object is formed clearly on the screen and measure the image distance and the object distance. Following that first trial, move the lens to a second position where the image is in focus (Do NOT move the screen or light source) and measure the image distance and the object distance. Then, move the screen about 10 centimeters towards the object and repeat the steps stated above for this position of the screen. Our goal was to obtain about 10 data points. After we have all our data, we were to share the data with the other groups on the Google Doc.

Part B:

To begin this part, you must set up the light source, secured at the 0 mm mark. Once again, the picture on the light source is the object. Before you adjust the lens or screen, you must measure the size of the arrow and record it as the object size. At this point, using the same lens as above, slide it 500 mm away (for our group, it was 750 mm) from the light source and move the screen until you get a focused image on the screen. Next, record the image distance and size of the image but be sure to remember that inverted images have negative heights. Finally, continue to move the lens or the screen to complete the remaining trials listed on the data table. Make sure to note that while doing some of these trials, they may produce no image. "None" may be a valid observation.



//Table 1//
 * Data/Observations:**
 * Part A**



//Table 2//
 * Part B**



//Table 3// //*//The table is in 4 significant figures.
 * There are many "none"s for both trial 4 and 5 because the image did not show up due to the rules of lenses.


 * Analysis/Calculations:**
 * Part A**

__Calculation for f(x intercept) and f(y intercept):__ (Repeat for y intercept as well)

__Calculation for percent difference between two values of focal length:__

__Calculation for f(average):__

__Calculation for M:__ (Repeat for Trials 2 and 3)
 * Part B**

__Calculation for percent difference between (1/do + 1/di) and 1/f:__ (Repeat for Trials 2 and 3)

__Calculation for percent difference between M and -di/do:__ (Repeat for Trials 2 and 3)


 * 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? The slopes of the various lenses all seem to be very similar. They all seem to be around -1. We can tell which lens has the bigger focal length by looking at the y-intercept of each graph. The bigger the y-intercept, the smaller the focal length will be. This is true based on a calculation that we had to perform earlier in the lab. This calculation showed that the y-intercept and focal length are inversely related therefore, as the y-intercept gets bigger, the focal length will be smaller.

2. Are the images real or virtual? How do you know? The images were real because we were able to see them. In addition, the rule for convex lenses states that if the image shows up on the opposite side of the lens from the object, the image is then real. This is opposite the rule for convex mirrors.

3. In all trials, were the images formed by the lens erect or inverted? Explain why. The images formed by the lens were all inverted. This occurred because as long as the object distance is greater than the focal length, the rule states that the image will always be real and inverted. We can also be sure they were inverted because when we focused the image on the screen, it was always upside-down.

4. Explain why, for a given screen–object distance, there are two positions where the image is in focus. For any given convex lens, there are two locations where real images are formed: beyond 2F and between 2F and F. As a result, there will be two positions where the image will be in focus - one beyond 2F and one between 2F and F.

5. How did the image distance change as the object was brought closer to the lens? As the object distance decreased, the image distance increased and vice versa.

6. How did the image height change as the object was brought closer to the lens? As the object distance decreased, the image height increased, until the object distance equaled the focal distance of the lens. At that point, there was no image. When the object distance became less than the focal distance, the image became virtual, and we could not see the image because the image was behind the object and the light rays never intersected.

7. Compare and contrast characteristics and images of lenses and mirrors. Image characteristics of lenses contrast those of mirrors. Image characteristics of concave mirrors are the same as image characteristics of convex lenses and vice versa. Like a convex mirror, concave lenses always produce virtual, reduced, and upright images. Like a concave mirror, convex lenses will produce either real or virtual images depending on object distance. The image can be enlarged or reduced, upright or inverted, again depending on object distance. One key difference between lenses and mirrors is sign notation. Unlike mirrors, the area through the lens, opposite the object, is positive and the area before the lens, same side of the object is negative. In mirrors, the opposite is true. Also, real images in lenses are opposite the object on the other side of the lens and virtual images are on the same side as the object. Again, the opposite is true of mirrors.

8. Explain your results for trials 4 and 5 in Part B. When we moved the object at the focal point and between the lens and focal point, no image was produced that we could see. There was no image between the focal length and lens because the image produced is actually virtual and cannot be seen in this experiment. Why there was no image for when the object was placed at the focal point can be proven mathematically using our equation. If the focal length and distance object is the same, 1/f and 1/do must be the same. Thus, 1/di must be 0 and no image can be produced.

9. For trial 5 in Part B, calculate the theoretical di, hi, and M. Show your work clearly.

10. Can you think of a way to project the image produced by an object between F and V onto a screen? No. The image produced when the object is between F and V would be virtual and it is impossible to project a virtual image on a screen because the light rays never actually intersect at that point.


 * Conclusion:**

In Part A, we observed many different image and object distances from our setup. We plotted these points on the graph after taking their inverse and found the trendline. The x and y intercepts of our trendline produced the inverse of our theoretical focal length. Our two values were relatively close to one another and had a percent difference of 0.78% and the average of the two was 49.31 cm. This is all evidence that proves the lenses equation to be true. Because our data produced great results that followed a trendline (with a .99 r squared) and our x and y intercepts were very similar, the fact needs to hold true that the sum of the inverse of object and image distance need to equal the inverse of the focal length. Of course there was a bit of error because our line was not perfect and this probably was due to our measurements. Our observations were based off of eye measurements and we could have definitely been a few millimeters off when measuring a distance. On top of this, it was tough to determine when the object was perfectly focused. This could have messed up our distances and ultimately caused some error. To address these sources of error, we would need to use a better measuring device that is a bit more accurate and have someone with perfect vision able to tell when the image was exactly focused.

In Part B, we placed the object at certain locations relating to the positioning of the lens and its focal point. In the first trial, we placed it at a distance more than two times the focal length. This produced an image that was reduced in size and inverted. I can conclude it was reduced because the image size was smaller than the object size. It was inverted because the image was shown to be "upside down." This was denoted using a negative sign. When the object was at two times the focal length, the image was inverted, but had the same size. Our observations showed the image size to be 1 mm off from the object size, but this was due to error. Between two times the focal length and the focal length, the image was still inverted, but was now bigger than the object. The magnification is what allowed me to conclude whether the object was enlarged, reduced, or the same size. There were two ways to calculate the magnification as seen in our tables and calculations. A magnification under 1 indicates that the image was reduced, 1 means it was the same size, and over 1 meant that it was enlarged. When we moved the object at the focal point and between the lens and focal point, no image was produced that we could see. There was no image between the focal length and lens because the image produced is actually virtual and cannot be seen in this experiment. Why there was no image for when the object was placed at the focal point can be proven mathematically using our equation. If the focal length and distance object is the same, 1/f and 1/do must be the same. Thus, 1/di must be 0 and no image can be produced. Through these observations it can be seen that our hypothesis was proven correct and these image characteristics and certain points do exist. There was no error in our qualitative observations for Part B, but there were quantitative errors as talked about above. They were probably due to the same errors talked about in Part A - inaccurate measurements and a difficult time determining when the image was entirely focused. Lens have many real life applications, most notably in glasses. These measurements are vital for uses of lens and it helps determine what an image will look like when they are in use. To sum this up, we found both hypothesis to be true - the lens equation works and there are certain image characteristics for an object placed at different locations.