Question 1: Absorption
At any given time, the number of photons inputted into the cavity must be equal to the number that have passed through the cavity without exciting an atom plus the number still in the cavity plus the number of excited atoms. Verify this conservation law by stopping the simulation and counting photons.
This simulation proves that the conservation law holds true. On this snapshot we see that there are 12 input photons. There are 9 excited atoms and 3 still in the cavity, that is a total of 12 photons.
Question 2: Direction of Spontaneous
During spontaneous emission, does there appear to be a preferred direction in which the photons are emitted?
No, they are emitted at random directions.
Question 3: Lifetime of Excited State
Does there appear to be a constant amount of time in which an atom remains in its excited state?
No, it seems like there is not constant time. They can be emitted at any moment.
Question 4: Stimulated Emission
Carefully describe what happens when a photon interacts with an excited atom. Pay careful attention to the phase and direction of the subsequent photons. (Can you see why this is called stimulated emission?)
As a photon interacts with an excited atom it stimulates the excited atom and a photon is emitted. More photons can be emitted from other excited atoms as these photons keep interacting.
Question 5: Pumping
Approximately what pumping level is required to achieve a population inversion? Remember, a population inversion is when the number of atoms in the excited state is at least as great as the number of atoms in the ground state.
On the simulation we see that pumping level should be about 10 to achieve a population inversion.
Question 6: Photon Emission
Although most photons are emitted toward the right in the simulation, occasionally one is emitted in another direction. Are the photons emitted at odd directions the result of stimulated or spontaneous emission?
The simulation shows that the photons emitted at odd directions are the result of spontaneous emission.
Thursday, June 14, 2012
Measuring a human hair
The purpose of this project is to measure accurately the thickness of a human hair by using diffraction.
We will obtain a human hair and tape it across the hole of a 3x5 note card. The card will be clamped parallel to the board several meter from the board. We will then shine a laser to the hair which will produce a diffraction pattern on the board. The first - order (m = 1) minima will show on the board and then we can use the diffraction formula to calculate the width.
Lenses
The purpose for this lab is to observe some characteristics of a converging lens when an object is placed on one side of the lens and the real, inverted image is placed on the other side of the lens.
We prepared the apparatus by placing the lens in the lens holder on the table, and then we placed the filament about 4 focal length distances from the lens. We then moved the screen back in forth to view an image and recorded our observation.
The focal length of the lens was printed on the box, which it states its f = 5cm.
-distance from center of lens to filament: d_o = 20cm
-distance from center of lens to image: d_i = 7.3cm
-height of filament: h_o = 9cm
-height of image : h_i = 4.2cm
-Magnification: M = h_i/h_o = 0.45
The image was inverted and real
We then reversed the lens and observed that there is no difference.
We moved the lens about 2 focal distances from the filament and record our measurements.
-distance from center of lens to filament: d_o = 10cm
-distance from center of lens to image: d_i = 10.7cm
-height of filament: h_o = 9cm
-height of image : h_i = 10cm
-Magnification: M = h_i/h_o = 1.11
The image was inverted and real
We moved the lens about 1.5 focal distances from the filament and record our measurements.
-distance from center of lens to filament: d_o = 7.5cm
-distance from center of lens to image: d_i = 15.7cm
-height of filament: h_o = 9cm
-height of image : h_i = 18.5cm
-Magnification: M = h_i/h_o = 2.1
The image was inverted and real.
I predict that the image would not be as bright if half the lens is covered with masking tape. I placed the masking tape and proved my prediction.
I then made a chart with all focal points measured and a few others.
When the lens is moved at a distance of .5 f, the image is not formed because it cannot focus. Its within the focal point. If viewing the lens from the inside the image can be viewed.
Slope = 0.451
y-int = 0.1547
We prepared the apparatus by placing the lens in the lens holder on the table, and then we placed the filament about 4 focal length distances from the lens. We then moved the screen back in forth to view an image and recorded our observation.
The focal length of the lens was printed on the box, which it states its f = 5cm.
-distance from center of lens to filament: d_o = 20cm
-distance from center of lens to image: d_i = 7.3cm
-height of filament: h_o = 9cm
-height of image : h_i = 4.2cm
-Magnification: M = h_i/h_o = 0.45
The image was inverted and real
We then reversed the lens and observed that there is no difference.
We moved the lens about 2 focal distances from the filament and record our measurements.
-distance from center of lens to filament: d_o = 10cm
-distance from center of lens to image: d_i = 10.7cm
-height of filament: h_o = 9cm
-height of image : h_i = 10cm
-Magnification: M = h_i/h_o = 1.11
The image was inverted and real
We moved the lens about 1.5 focal distances from the filament and record our measurements.
-distance from center of lens to filament: d_o = 7.5cm
-distance from center of lens to image: d_i = 15.7cm
-height of filament: h_o = 9cm
-height of image : h_i = 18.5cm
-Magnification: M = h_i/h_o = 2.1
The image was inverted and real.
I predict that the image would not be as bright if half the lens is covered with masking tape. I placed the masking tape and proved my prediction.
I then made a chart with all focal points measured and a few others.
When the lens is moved at a distance of .5 f, the image is not formed because it cannot focus. Its within the focal point. If viewing the lens from the inside the image can be viewed.
Slope = 0.451
y-int = 0.1547
Relativity of Time/Length
Relativity of Time:
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?
The distance traveled by the moving clock is greater than the distance traveled by the stationary clock
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?
The time interval for the stationary clock is slower than the time interval for the moving clock.
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?
The distance would appear to be the same.
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?
y = 1/(sqrt [1+(v^2/c^2)]), as velocity decreases the difference in time decreases.
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 (y) of 1.2.
Using the dilation formula, the predicted time is about 8 e-6 seconds.
6.If the time interval between departure and return of the light pulse is measured to be 7.45 e-6 seconds by an earth-bound observer, what is the Lorentz factor of the light clock as it moves relative to the earth?
Using the dilation formula, the Lorentz factor is about 1.12.
The effect happening in theses situations are know as "Time dilation." This effect happens when an event occurs and the event is viewed at two frame of references. One reference is at the same space point where the event occurs, this frame of reference is moving with a constant velocity relative to the a second frame of reference. The time interval from the second frame of reference is larger than the time interval from the frame of reference where the event occurred.
Relativity of Length:
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?
No, it does not matter.
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?
The product of the measured time interval on the light clock and the Lorentz factor is equal to the time inteval on earth. The time intervals will be the same as long as you take time dialation into account.
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?
Only if the Lorentz factor is equal to 1, this can happen if the velocity of the the moving light clock is relatively small compared to c.
4)A light clock is 1000m 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?
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?
The distance traveled by the moving clock is greater than the distance traveled by the stationary clock
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?
The time interval for the stationary clock is slower than the time interval for the moving clock.
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?
The distance would appear to be the same.
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?
y = 1/(sqrt [1+(v^2/c^2)]), as velocity decreases the difference in time decreases.
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 (y) of 1.2.
Using the dilation formula, the predicted time is about 8 e-6 seconds.
6.If the time interval between departure and return of the light pulse is measured to be 7.45 e-6 seconds by an earth-bound observer, what is the Lorentz factor of the light clock as it moves relative to the earth?
Using the dilation formula, the Lorentz factor is about 1.12.
The effect happening in theses situations are know as "Time dilation." This effect happens when an event occurs and the event is viewed at two frame of references. One reference is at the same space point where the event occurs, this frame of reference is moving with a constant velocity relative to the a second frame of reference. The time interval from the second frame of reference is larger than the time interval from the frame of reference where the event occurred.
Relativity of Length:
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?
No, it does not matter.
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?
The product of the measured time interval on the light clock and the Lorentz factor is equal to the time inteval on earth. The time intervals will be the same as long as you take time dialation into account.
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?
Only if the Lorentz factor is equal to 1, this can happen if the velocity of the the moving light clock is relatively small compared to c.
4)A light clock is 1000m 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?
The effect happening in this situations is know as "Length Contraction." This effect happens when a distance between two points is measured in the rest frame of reference, known as proper length, is moving at a constant velocity relative to a second frame of reference. The length of the two points measured at the second frame of reference is shorter than the proper length.
Monday, June 11, 2012
Introduction to Sound
For this experiment we observed the wave patterns created by sound. Using Logger Pro, we were able to view the patterns created.
First, a student from our group said "AAAAA" smoothly into the microphone and collected data. The wave patter created is shown below.
Here we see that this wave is periodic, the same pattern is created from the highest point to the next.
In the .03 seconds of data collected 3.5 waves were created.
The period of the wave was approximately .0085 seconds, which was determined by reading the time of one wavelength on LoggerPro.
The frequency was approximately 117.6 Hz, which is the reciprocal of the period.
f=1/T
The wavelength was 2.89m, which was calculated using a relationship between the speed of sound and frequency
wavelength (lambda) = speed of sound/frequency = v/f
The amplitude was determined by reading the length from the middle of the wave to its highest point, the aplitude for this wave was 1.8 (arbitrary).
If the duration of the sample was changed 10 times longer the wave properties will not change. The same pattern will appear but the only difference is that there will be 10 times more waves on the screen as shown below.
The experiment was done once more with a different student and the wave created is shown below.
Now, the same experiment was done using a tuning fork instead of a human voice.
Using the same tuning fork, we collected data for a sound that was not as loud.
Standing Sound Waves- open tube
For this experiment the professor spun a tube and recorded the sound wave created by the tube. Using our knowledge of waves and their properties we calculated the length of the tube using the data recorded.
Wednesday, April 11, 2012
Diffraction
The purpose of this lab is to determine the distance (d) between grooves on a CD by diffraction.
We first arrange a laser so it shines on the disk and as it hits the grooves on the disk it causes diffraction. We measure the distance of the light difrracted from the middle to the point of first order maxima and the length from the light source to the wall where the light showing.
Since it causes constructive interference at the first maxima we used the formula
We first arrange a laser so it shines on the disk and as it hits the grooves on the disk it causes diffraction. We measure the distance of the light difrracted from the middle to the point of first order maxima and the length from the light source to the wall where the light showing.
Since it causes constructive interference at the first maxima we used the formula
d*sin(theta) = m*lambda
and solve for d.
We see that the distance between grooves on a CD is about 1.5 micrometers.
Concave and Convex Mirrors
The pupose for this experiment is to observe the images formed by two different types of mirrors, concave and convex mirros. To analyze the characteristics of the images due to the curvature of the mirror, distance of the object to the mirror and reflected rays.
The first mirror we observed was a convex mirror.We first placed an object in front of the mirror, the image appeared smaller than the actual object. The image was upright and was located inside the mirror, farther away relative to the distance of the mirror and object.
As the object is moved closer to the mirror, the image gets larger almost to its actual size.
As the object is moved away from the mirror the image gets smaller.
The second mirror was a concave mirror.We first placed an object in front of the mirror the image appeared larger than the actual object and inverted, this is when the object was located behind the curvature point.
Once the object was moved closer the image was upright.
The image was located relatively closer than the mirror and object.
The magnification of the image is determined by using the lateral magnification:
m = y(i) / y(o) = -s(i) / s(o)
Where y(i) is the height of the image and y(o) is the height of the object.
s(i) is the distance from the mirror to the image and s(o) is the distance from the mirror to the object.
Tuesday, April 10, 2012
Standing Waves (short lab)
The purpose of this lab is to determine how frequency and wavelength are related.
First, we had two students hold each end of a spring. Then, we had the students create a standing wave by shaking the spring back and forth in the horizontal direction.
We first started with one wavelength and increased it by one for each trial, with a total of four trials. Two other students also time the amount it took for 20 waves and determined the frequency by dividing the total cycles by the amount of time.
First, we had two students hold each end of a spring. Then, we had the students create a standing wave by shaking the spring back and forth in the horizontal direction.
We first started with one wavelength and increased it by one for each trial, with a total of four trials. Two other students also time the amount it took for 20 waves and determined the frequency by dividing the total cycles by the amount of time.
This is the frequency that was determined:
Trial 1 : 2.35 Hz
Trial 2 : 2.86 Hz
Trial 3 : 1.67 Hz
After recording the data, it was then graphed and determined a relationship between wavelength and frequency. The wavelength and frequency seem to be inversely proportional to one another.
Saturday, March 31, 2012
Standing Waves
The goal for this experiment is to investigate and have a basic understanding of force driven standing waves.
A standing wave is a transverse wave traveling along a medium, in this case a string, and then reflected back and returns interfering with other waves. When interfering, at resonance, a standing wave is created with nodes and antinodes in the wave pattern.
Equations:
-Transverse wave in the positive x direction.
y1 = A sin(kx - wt),
where k = 2*pi/lambda & w = 2*pi*f
-Transverse wave reflected from fixed end
y2 = A sin(kx + wt)
-At resonance, the two waves combine and become the some of both waves.
y = y1 + y2 = (2A sin (kx))*cos(wt)
Other equations:
-Wavelength for each harmonic of vibrating string.
Lambda = 2L/n
-If frequency is known, the wave speed can be determined with fundamental wave velocity equation.
v = f*lambda
-Substituting both.
f = vn/2L
-velocity of wave traveling on a string.
v = sqrt(T/mu),
where T is tension & and mu is mass per unity length.
Experiment:
We used mechanism that drives a wave on one side of the string, and the other side of the string was pulled by a hanging mass creating tension. We adjusted the frequency of the driving mechanism to find each harmonic and measured the wavelength.
For case 1 we had a string of length 140 cm and a hanging mass of 200 g.
-Data:
16.5 Hz, 1 loop, 2 nodes, lambda 140 cm
32.5 Hz, 2 loops, 3 nodes, lambda 67 cm
45.6 Hz, 3 loops, 4 nodes, lambda 45 cm
56.7 Hz, 4 loops, 5 nodes, lambda 37 cm
74.6 Hz, 5 loops, 6 nodes, lambda 28 cm
87.4 Hz, 6 loops, 7 nodes, lambda 23 cm
For case 2 we had a string of length 186.5 cm and a hanging mass of 50 g.
Wednesday, March 7, 2012
Fluid Dynamics
In this experiment we filled a bucket with water and recorded the time elapsed as the water drained from a small hole at the bottom of the bucket. Then, calculated the theoretical time elapsed and compared both results.
Experiment: The bucket was filled with water 2.3 inches above the small hole, the hole had a diameter of approximately 0.6 centimeters. We ran six trials and recorded the time it took for 473 mL (16 ounces) of water to flow out of the bucket.
Volume emptied (V) = 16 ounces = 473 mL
Height of water (h) = 2.3 in = 0.192 feet
Area of drain hole (A) = pi*(.3cm)^2 = .28 cm^2 = 0.0003014 ft^2
Acceleration due to gravity (g) = 32 ft/s^2
Trial # time (t_actual)
1 25.06 s
2 24.85 s
3 25.24 s
4 25.36 s
5 25.17 s
6 26.09 s
R = V/t = Av & v = sqr(2gh) -----> V/t = A[sqr(2gh)] ----> t(theoretical) = V/A[sqr(2gh)]
Experiment: The bucket was filled with water 2.3 inches above the small hole, the hole had a diameter of approximately 0.6 centimeters. We ran six trials and recorded the time it took for 473 mL (16 ounces) of water to flow out of the bucket.
Volume emptied (V) = 16 ounces = 473 mL
Height of water (h) = 2.3 in = 0.192 feet
Area of drain hole (A) = pi*(.3cm)^2 = .28 cm^2 = 0.0003014 ft^2
Acceleration due to gravity (g) = 32 ft/s^2
Trial # time (t_actual)
1 25.06 s
2 24.85 s
3 25.24 s
4 25.36 s
5 25.17 s
6 26.09 s
The average of the six trials was 25.30 seconds. We then calculated the theoretical time by using the Flow rate equation, Continuity equation and Torricellis Theorem, which relates the speed of fluid flowing out an opening to the height of fluid above the opening.
t (theoretical) = 15.14 s
Calculate Error: |theoretical - actual|/theoretical x 100 = % error
|25.3s - 15.14s|/15.14 x 100 = 67% error
Comparing our measured value with the theoretical value gave us a 67% error. This does not agree within uncertainty. We then assumed that the measured diameter was inaccurate and decided to solve for the actual diameter using the same equations and the data recorded.
v = sqr(2*32*0.192) = 3.505 ft/s, R = V/t = (0.0160 ft^3)/25.30s = 6.324*10^-4 ft^3/s
V/t = A[sqr(2gh)] --- > 6.324*10^-4 ft^3/s = pi*r^2*3.505
r = 0.0076 ft = 0.28 cm, diameter = 0.56 cm
|0.6cm - 0.56cm|/0.6 x 100 = 6.67 % error
Here we see that our diameter should had been 0.56 cm. To check our result we compared both the calculated and measured diameters, this gave us a percent error of 6.67% which is within our uncertainty. This proves that our measurements were correct but there is still an unknown source that altered our results drastically.
There may be a few reasons why our result for our time had a big error. It is possible that we did not measured the height of the water correctly since we did not have a good angle reading the measurement while looking into the bucket. The increments on the beaker were not small enough to read exactly 473 mL, it was an approximate. Having better measuring equipment would had definitely lowered the error on our results.
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