Monday, December 12, 2011
17.2 Modern Physics: Relativity of Length
In this online activity we learned how relativity distorts the length of objects as they approach the speed of light. We will be using length contraction equation L=L_0/sqrt[1-(v^2/c^2)]. As in the previous activity questions will be followed with a screenshot and the appropriate answers.
Question #1: Imagine riding on the left end of the light clock. A pulse of light departs the left end, travels to the right end, reflects, and returns to the left end of the light clock. Does your measurement of this round-trip time interval depend on whether the light clock is moving or stationary relative to the earth?
Answer: The round trip time depends on whether the clock is stationary or moving.
Question#2: Will the round-trip time interval for the light pulse as measured on the earth be longer, shorter, or the same as the time interval measured on the light clock?
Answer: The round-trip time will be measured to be shorter than the time measured relative to the earth.
Question #3: You have probably noticed that the length of the moving light clock is smaller than the length of the stationary light clock. Could the round-trip time interval as measured on the earth be equal to the product of the Lorentz factor and the proper time interval if the moving light clock were the same size as the stationary light clock?
Answer: Yes, if the loreanzt factor is known and the time required for a moving clock to travel a given distance we can use the formula stated at the beginning of the activity to calculate the time required to reach the same distance relative to the stationary frame.
Question #4: A light clock is 1000 m long when measured at rest. How long would earth-bound observer's measure the clock to be if it had a Lorentz factor of 1.3 relative to the earth?
Answer: The Lorentz factor is 1.3, and the length is 1000m at rest, so we use the formula given at the beginning of this activity to calculate the contracted length. We obtain 1000/1.3 = 769 meters.
Question #1: Imagine riding on the left end of the light clock. A pulse of light departs the left end, travels to the right end, reflects, and returns to the left end of the light clock. Does your measurement of this round-trip time interval depend on whether the light clock is moving or stationary relative to the earth?
Question#2: Will the round-trip time interval for the light pulse as measured on the earth be longer, shorter, or the same as the time interval measured on the light clock?
Answer: The round-trip time will be measured to be shorter than the time measured relative to the earth.
Question #3: You have probably noticed that the length of the moving light clock is smaller than the length of the stationary light clock. Could the round-trip time interval as measured on the earth be equal to the product of the Lorentz factor and the proper time interval if the moving light clock were the same size as the stationary light clock?
Question #4: A light clock is 1000 m long when measured at rest. How long would earth-bound observer's measure the clock to be if it had a Lorentz factor of 1.3 relative to the earth?
17.1 Modern Physics: Relativity of Time
This online activity was used to help us understand the basic concepts of relativity. In these problems we will explore how space and time are distorted as different frames approach the speed of light. We will make use of the time dilation equation which is (delta)t' = (delta)t/sqrt[1-(v^2/c^2)]. Screenshots of each of the question a posted below with their corresponding answers.
Question#1: How does the distance traveled by the light pulse on the moving light clock compare to the distance traveled by the light pulse on the stationary light clock?
Answer: The distance traveled by the moving light clock is greater than the distance travel by the stationary light clock.
Question #2: Given that the speed of the light pulse is independent of the speed of the light clock, how does the time interval for the light pulse to travel to the top mirror and back on the moving light clock compare to on the stationary light clock?
Answer: Since the speed of light is constant, the moving light clock must travel a further distance and therefore must take more time to complete one cycle.
Question #3: Imagine yourself riding on the light clock. In your frame of reference, does the light pulse travel a larger distance when the clock is moving, and hence require a larger time interval to complete a single round trip?
Answer: In the moving frame of reference, the light pulse does not travel a greater distance, thus the time required for the light to make one cycle in the moving frame is the same time as the light to make one cycle in the stationary frame.
Question #4: Will the difference in light pulse travel time between the earth's timers and the light clock's timers increase, decrease, or stay the same as the velocity of the light clock is decreased?
Answer: If the velocity of the light clock is decreased, the difference of earth's timer and the light clock's timers also decreases.
Question #5: Using the time dilation formula, predict how long it will take for the light pulse to travel back and forth between mirrors, as measured by an earth-bound observer, when the light clock has a Lorentz factor (γ) of 1.2.
Answer: The time dilation of a moving clock with a Lorentz factor of 1.2 will have a time 1.2 times that of the original time.
Question #6: If the time interval between departure and return of the light pulse is measured to be 7.45 µs by an earth-bound observer, what is the Lorentz factor of the light clock as it moves relative to the earth?
Answer: The measured time is 7.45 µs, we used the simulator and found that the Lorentz factor is around 1.12, which is what about the same number we calculated.
Question#1: How does the distance traveled by the light pulse on the moving light clock compare to the distance traveled by the light pulse on the stationary light clock?
Question #2: Given that the speed of the light pulse is independent of the speed of the light clock, how does the time interval for the light pulse to travel to the top mirror and back on the moving light clock compare to on the stationary light clock?
Question #3: Imagine yourself riding on the light clock. In your frame of reference, does the light pulse travel a larger distance when the clock is moving, and hence require a larger time interval to complete a single round trip?
Answer: In the moving frame of reference, the light pulse does not travel a greater distance, thus the time required for the light to make one cycle in the moving frame is the same time as the light to make one cycle in the stationary frame.
Question #4: Will the difference in light pulse travel time between the earth's timers and the light clock's timers increase, decrease, or stay the same as the velocity of the light clock is decreased?
Question #5: Using the time dilation formula, predict how long it will take for the light pulse to travel back and forth between mirrors, as measured by an earth-bound observer, when the light clock has a Lorentz factor (γ) of 1.2.
Question #6: If the time interval between departure and return of the light pulse is measured to be 7.45 µs by an earth-bound observer, what is the Lorentz factor of the light clock as it moves relative to the earth?
Tuesday, October 18, 2011
Experiment 12: CD Diffraction
In this experiment we used a laser with a known wavelength to measure the width of the grooves in a compact disk. To do this we first set up the experiment with a laser at an angle of about 45 degrees to the normal of the surface of the disk. As the laser became hit the surface of the CD the laser beam would diffract and split into multiple beams of light. The reason why we decided to place the CD at a 45 degree angle was so the the diffracted beams could be shown a white board that was facing parallel to the laser beam before hitting the CD. The picture below better illustrates the apperatus.
Using this setup we measured the distance from the CD to the white board to be 0.070 meters +/- 0.0005. We also know that the wavelength of the laser to be 670 nm +/- 0.10 and that the distance from the central maximum to the first maximum is 0.028 m +/- 0.0005. Using the equation y=[m*(lambda)*x]/a where x = 0.7 meters, lambda = 670*10^-9 meters, m = 1, and y = 0.028 meters we can solve for (a) (the slit spacing in the CD). As a result we found that the CD was 1.675*10^-6 meters. The accepted value for the slit spacing is 1.6*10^-6 meters which gives us a percent difference of 4.6%. This falls well within our percent uncertainty of 31%.
This experiment allowed us to find a real world application for laser diffraction and how it can be used to measure defects in CD spacing. With more precise instrumentation would could significantly decrease our percent uncertainty and difference to more accurately the CD's spacing. Also, in this experiment we made a small angle approximation in order to use our equation y=[m*(lambda)*x]/a. Uncertainty due to this approximation was unaccounted for but because of the small angles between the central and first maximums this uncertainty could be safely ignored.
Using this setup we measured the distance from the CD to the white board to be 0.070 meters +/- 0.0005. We also know that the wavelength of the laser to be 670 nm +/- 0.10 and that the distance from the central maximum to the first maximum is 0.028 m +/- 0.0005. Using the equation y=[m*(lambda)*x]/a where x = 0.7 meters, lambda = 670*10^-9 meters, m = 1, and y = 0.028 meters we can solve for (a) (the slit spacing in the CD). As a result we found that the CD was 1.675*10^-6 meters. The accepted value for the slit spacing is 1.6*10^-6 meters which gives us a percent difference of 4.6%. This falls well within our percent uncertainty of 31%.
This experiment allowed us to find a real world application for laser diffraction and how it can be used to measure defects in CD spacing. With more precise instrumentation would could significantly decrease our percent uncertainty and difference to more accurately the CD's spacing. Also, in this experiment we made a small angle approximation in order to use our equation y=[m*(lambda)*x]/a. Uncertainty due to this approximation was unaccounted for but because of the small angles between the central and first maximums this uncertainty could be safely ignored.
Tuesday, October 11, 2011
Experiment 11: Measuring a Human Hair
In this experiment we used the a micrometer and a laser to measure and compare the thickness of a human hair. We began by using the laser and pointing it directly at the hair. We then measured the diffraction of light, the distance our image was from the object, and knew that the wavelength of our laser to be about 670 nm +/- 20 nm.
We then took three trials and on each trial we measured the distance from the hair to the image on a board, the node number, y_m, and lamda which we took to be 670 nm +/- 20nm. We then used the equation d=L*m*(lamda)/y to find the thickness of the hair d.
We then took three trials and on each trial we measured the distance from the hair to the image on a board, the node number, y_m, and lamda which we took to be 670 nm +/- 20nm. We then used the equation d=L*m*(lamda)/y to find the thickness of the hair d.
Trial | Length (L, m) | Node Number (m) | y (m) | Lamda (m, +/- 20 nm) | d (m) |
1 | 1.68 | 4 | 0.063 | 6.7E-07 | 7.15E-05 |
2 | 2 | 4 | 0.076 | 6.7E-07 | 7.05E-05 |
3 | 3.62 | 3 | 0.11 | 6.7E-07 | 6.61E-05 |
We found after three trails the thickness of the hair to be about 6.9*10^(-5) meters thick. This fits within our estimation that the thickness of the hair would be between 1.7*10^(-5) to 1.8*10^(-4) meters thick. Using the micrometer we found the thickness of the hair to be about 7*10^(-5) meters which falls well within our trials for the thickness of the hair.
Experiment 10: Lenses
We began this experiment by measuring the focal distance of the lens that were were using. To do this we used the sun as our source and moved the lens until the light was focused at a single point. We then measured the distance from this point to the lens. We determined that the focal distance for our particular lens to be about 19.5 centimeters. We then shined an image through our lens and measured the object distance, image distance, object height, image height, and we calculated the magnification. We also stated whether or not the image was inverted.
We found that in every case the image was inverted, but when be placed the object within the focal distance no image could be seen. Only when you looked though the lens could you see an image in the lens. To describe the relationship between image distance and object distance we took the inverse of d_i and plotted it against the negative inverse of d_o. We then obtain the following graph.
This graph illustrates that the equation 1/d_o + 1/d_i = 1/f. However this graph shows that the focal length of our lens to actually be 20.9 cm long. This is a percent difference of 6.6% which is in a reasonable percent error when considering the error that went into this experiment. If we wished to decrease the percent error we could have used more accurate measuring equipment and have used a different method when it came to measuring the focal length of our lens.
Object distance (d_o, cm) | Image distance (d_i, cm) | Object height (h_o, cm) | Image height (h_i, cm) | Magnification | Type of Image |
100 | 26.5 | 3 | 0.8 | 0.266 | Inverted |
80 | 28.5 | 3 | 1 | 0.333 | Inverted |
60 | 32 | 3 | 1.5 | 0.5 | Inverted |
40 | 43.5 | 3 | 3.5 | 1.16 | Inverted |
30 | 71 | 3 | 6.7 | 2.23 | Inverted |
We found that in every case the image was inverted, but when be placed the object within the focal distance no image could be seen. Only when you looked though the lens could you see an image in the lens. To describe the relationship between image distance and object distance we took the inverse of d_i and plotted it against the negative inverse of d_o. We then obtain the following graph.
This graph illustrates that the equation 1/d_o + 1/d_i = 1/f. However this graph shows that the focal length of our lens to actually be 20.9 cm long. This is a percent difference of 6.6% which is in a reasonable percent error when considering the error that went into this experiment. If we wished to decrease the percent error we could have used more accurate measuring equipment and have used a different method when it came to measuring the focal length of our lens.
Tuesday, October 4, 2011
Experiment 9: Concave and Convex Mirrors
In this experiment we gained an understanding of how concave and convex mirrors create an image. We began by looking into these two types of mirrors and analyzed the differences between them. After that we used simple geometry to draw diagrams that would help illustrate how each of the mirrors formed their images. The first diagram shows the image from a convex mirror.
The second diagram shows how we found the image in a concave mirror.
To create each of these diagram we began by drawing a line from the object to the center. Then we drew a line from the object parallel to the optic axis and then when the line hit the mirror we drew the line though the focus. Next we drew a line to the focus but when the line hit the mirror we then drew it parallel to the optic axis. Where all these lines intersected was our image. From these diagrams we could easily see where our image was relative to our object and out mirror and whether the image was erect or inverted. We were also able to tell that the image would have some magnification.
The second diagram shows how we found the image in a concave mirror.
To create each of these diagram we began by drawing a line from the object to the center. Then we drew a line from the object parallel to the optic axis and then when the line hit the mirror we drew the line though the focus. Next we drew a line to the focus but when the line hit the mirror we then drew it parallel to the optic axis. Where all these lines intersected was our image. From these diagrams we could easily see where our image was relative to our object and out mirror and whether the image was erect or inverted. We were also able to tell that the image would have some magnification.
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