Wednesday, June 13, 2012

Determining Planck's Constant

In this experiment we were using LEDs (Light Emitting Diodes) to determine Planck's Constant experimentally. We tested two light emitting diodes which were blue and yellow. They shine at different levels of intensity at different voltages because of the color of light that they emit. The values for the wavelength(λ) of blue and yellow light were estimated. The picture below is of the an LED with a diffraction grating in front of the lens. The only color seen through the diffraction grating is a deep blue color and a pale yellow color.






The energy was determined by the voltage output coming from the power supply. When the electrons pass through a potential,  they will have an energy in terms of electron volts(eV).
E_Blue = 1.95 eV λ_Blue = 450 nm
E_Yellow = 2.76 eV λ_Yellow = 590 nm

The experimental value of Planck's constant can be determined by plotting the values of λ vs 1/E and using the slope. The slope can be used to calculate Planck's constant.


 



The slope from the graph is m = 1.490225*10^-25.  h can be found through this equation, 
m = hc, where m is the slope and it is a relation between the wavelength and the energy, h is Planck's constant, and c is the speed of light.  m/c = h 
h = 4.967 * 10^-34 Js is the calculated value. The percent error of this value compared to the actual value is  25%. We have this large error because only two colors were used to determine the slope and the values of the wavelength were estimated instead of calculated. 



Light and Matter Waves

When light and matter waves pass through an object or an opening, they diffract. The intensity can be calculated with Huygens' Principle along with the ideas of the superposition principle. These different ideas will be simulated with python(x,y). These simulations will help us visualize how the waves behave. We will begin by observing how the electric field from a point source changes with a detector. The detector will detect through all of the points in the space being observed.


When the slit space(y) is 12mm and a wavelength(λ) of 2mm and there is one point source the will be seen in this form.
y = 12 mm, λ = 4 mm 1 point source










y = 12 mm, λ = 8 mm 1 point source









y = 12 mm, λ = 2 mm 2 point sources









y = 12 mm, λ = 4 mm 2 point sources









y = 24 mm, λ =4 mm 2 point sources









y = 12 mm, λ = 8mm 2 point sources 












y = 24 mm, λ = 8mm 2 point sources 








In the next set of pictures we changed the amount of sources and the number of detectors in the wave.

There are 50 sources (s) and 500 detectors (d)



This is when M = 5 and N = 7


This is when M = 50 and N = 500


This is when M = 51, N = 500, the slit distance = 1μm and the slit width = 5μm


This is when M = 51, N = 600, the slit distance = 0.5 μm and the slit width = 10μm 


This is when M = 51, N = 600, the slit distance is 50mm and the slit width is 25μm


This is when the length of the wave is 600nm


This is when the length of the wave is 700nm


This is when the screen distance is 5 mm



This is when the screen width is 50 mm 


This is when the screen width is 5 mm


There are different lines of code that help us program these different simulations. The code below helps us calculate our coordinate system that we used. 
N = 100
dX = screendist/N
dY = screendist/N
Xcoords = arange(0, screendist+2*dX, dX)
Ycoords = arange(-screendist/2, screendist/2 + 2*dY, dY)
[xd,yd] = meshgrid(Xcoord,Ycoord)


When we are graphing 2D images of our 3D graphs, it gives us a representation of how the graph behaves and any differential area. It is easier visually, to interpret a 2D graph versus a 3D graph. Some of graphs go to a zero point because this is when destructive interference occurs at its most intense amount. 
The maximums of the graphs change as you go away from the center. The further the detector is away, the harder it is to detect the waves. When the detector is close it can tell us a lot of information of how the waves behave. 










Wednesday, May 30, 2012

CD Diffraction

In this lab we began by shining a laser on the CD and looking at the diffraction of the laser onto another surface. The observed points would give us the first maxima of the laser. We then measured the distance we could determine the diffraction grating of the CD. The wavelength of the laser was already known for this lab.

The laser on the CD



The diffraction of the laser from the CD

We can find the diffraction grating with this formula dsin(θ) = mλ, in this formula d represents the diffraction grating. We can rearrange this equation for d and the result is, d = mλ/sin(θ). Sin(θ) is the distance measured from the maxima. m is an integer value based on how many maximas have been measured. λ is the wavelength of the laser. L and x are the sides of the triangle used for sin(θ).Arctan(x/L) can be used to find the value of θ

L = 68.5 cm + 0.2 cm
x = 28.8 cm + 0.2 cm
The wavelength of the laser was  λ  = 632.8 nm. 
The value of θ = 22.80° + 0.00269°
m = 1 because we measured the distance to the first maxima. 

When plugging these values into the formula we can determine that d = 1.633 * 10^-6m = 1633nm + 1.8735 * 10^-6 nm
The actual value of the CD's diffraction grating is 1600 nm. The percent error for this lab is 2.1%. The error from this lab is low compared to the usual value of 5%. 
The uncertainty for the values of L and x came from our inaccuracy in the measuring tools. 
The error for terms that were not measured during the lab were determined by using partial derivatives.
There is no error for λ because the laser mas made with extreme accuracy and precision. 
The error for θ can be found by:
 Δθ = sqrt((θ/x)^2*Δx^2 + (θ/L)^2*ΔL^2) = 0.00269
The error for d can be found by:
Δd = sqrt((∂d/θ)^2*Δθ^2+(∂d/λ)^2*Δλ^2), Δλ is zero because there is no error and the error reduces to: d = sqrt((∂d/θ)^2*Δθ^2) = 1.8735*10*-15 m




Color And Spectra

In this lab we began by setting an apparatus to observe the different colors from the various gases. We used two meter sticks, a diffraction grating, and a light source. We would observe the different light patterns and had a partner mark down the different colors.
 We had the points when the color changed marked down on the meter stick next to the metal water bottle, while the observation point was at the end of the other meter stick. The possible colors that would appear were in the order violet, blue, green, yellow, orange, red which are in order from shortest wavelength to longest wavelength.




The first light that we observed was white light. When the light is first observed a string of rainbow colors are instantly visible. The closest color to the bulb was blue while the furthest color from the bulb was red. The colors appeared in the order of violet, blue, green, yellow, red.












The distance from the light source (D) was measured and is listed below.
Violet 19.9+ 0.2 cm, Blue 22.3 + 0.2cm, Green 25.1 + 0.2 cm Yellow 26.9 + 0.2 cm, Red 41.0 + 0.2cm.
The diffraction grating (d) was 500 lines per mm. The distance to the slit (L) was 1m + .01 m, with this data the corresponding wavelengths can be determined.
 λ = (D*d)/sqrt(L^2 + D^2)
λ_ Violet = 390nm + 9nm, λ_Blue = 435nm + 9nm, λ_Green =  487nm + 9nm, λ_Yellow = 520nm + 9nm, λ_Red = 759nm + 9nm,
In order to get an accurate result, we mapped our measured values to a graph of actual values to see how our values compare to the actual result. We used this formula, λ = mλ' + λ_0. λ is our adjusted wavelength compared to the actual scale. In this formula λ' is our experimental λ and λ_0 is a constant that will correct the experimental λ to the scale of the actual λ. The slope was obtained by comparing values from the actual λ and the experimental λ. The constant λ_0 can be found from the equation, λ - λ _1 = m(λ'-λ_1'), λ_1 prime is an actual value while λ_1' is an experimental value. After some algebra, the equation results to λ = mλ' - mλ_1'+ λ_1, the quantity (-mλ_1' + λ_1) = λ_0.
Our λ values mapped to the actual scale can be determined from this.
λ mapped: λ_Violet = 466nm + 44nm, λ_Blue = 496nm + 44nm, λ_Green = 531nm + 44nm, λ_Yellow = 553nm + 44nm, λ_Red =716nm + 44nm.
The uncertainty for these measurements were determined by finding the difference  between of the averages of the observed value from the actual value.
In the next part of the experiment we observed the light from hydrogen gas. Only four different colors were observed from the spectra.

The observed values (D) from the light source were:
Violet = 22.1 + 0.2cm Blue = 25.2 + 0.2cm Yellow = 30.8 + 0.2cm Red = 35.0 + 0.2cm

The calculated λs from this were:
λ_Violet = 431nm + 24nm, λ_Blue = 489nm + 24nm, λ_Yellow = 589nm + 24nm, λ_Red = 661nm + 24nm.

Our λ values mapped to the actual values are:
λ_Violet = 391 nm + 42nm, λ_Blue = 435 nm + 42nm, λ_Yellow = 512 nm + 42nm, λ_Red = 567nm + 42nm.

Lastly, we observed the line spectra for an unknown gas and had to determine which gas it was.



The observed values (D) from the light source were:
Violet = 20.1 + 0.2cm Blue = 22.7 + 0.2cm Green = 26.1 + 0.2cm Yellow = 29.9 + 0.2cm Red = 35.8 + 0.2cm







The mapped values of λ are:
λ_Violet = 392 nm + 8 nm λ_Blue = 439 nm + 8 nm λ_Green = 498 nm + 8 nm λ_Yellow = 562 nm + 8 nm λ_Red = 658 nm + 8 nm. From the following data we can determine that our unknown gas was helium.


In this lab we were able to determine the spectral lines when we looked at light, or gases through a diffraction grating. The distance of the different colors observed we were able to map it to the actual values to determine how accurate we were.



Monday, May 28, 2012

ActivPhysics (The Laser)

In this Activphysics we did problems regarding how lasers work

Question 1: 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 formula can describe the behavior of the photons where N represents photon, N_in = N_Out + N_Miss + N_Excited.
In the picture below it corresponds to 12 = 0 + 3 + 9


Question 2: During spontaneous emission, does there appear to be a preferred direction in which the photons are emitted? 
The photons leave in any direction, they do leave in a preferred direction. 

Question 3: Does there appear to be a constant amount of time in which an atom remains in its excited state?
No, The photons are not excited for a certain amount of time, they will leave at random times.

Question 4: 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?)
The photons will collect energy from the excited atoms, this effect will continue if the photon hits more excited atoms. 

Question 5: 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. 
A pumping level of about 75 is enough for population inversion to occur.

Question 6: 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?
It is the result of spontaneous emission.





ActivPhysics(Relativity)

In this ActivPhysics lab  we answered various questions regarding relativity. These questions concentrated on time dilation and length contraction.

Time Dilation

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?
The light on the moving clock traveled a longer distance so it will have a longer time on it. The stationary clock travels a shorter distance.

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?
The time on the light moving in a diagonal will have a a longer time on it because it is traveling a longer distance. The clock on the stationary platform will have a shorter time because it is traveling a shorter distance. 

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?
The light pulse on the moving platform does travel a longer distance, however, in my frame of reference it takes me the same amount of time as the stationary clock. The light appears to be moving stationary to me so the time is same. Relative to the stationary clock, it will record a longer time.
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?
The difference in the timers will increase if the length of the light pulse traveled increases and decrease if the path traveled 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.
The normal time that it takes to travel is 6.67μs, with Lorentz factor at 1.2 it will take 1.2* 6.67μs to travel the distance. This results in 8.00μs. 

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?
γ*6.67μ = 7.45μs -> 7.45μs/6.67μs = γ Gamma results in 1.116 which is approximately 1.12.

Length Contraction

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?
If I am on the moving clock, the recorded time will be the same to me as the stationary clock. The light will appear to move the same distance that  I do. In my frame of reference, The clock doesn't move. 

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? 
The round trip time interval will be longer because it travels a longer distance. The time observed on both frames will be the same. The time measured on the moving frame relative to the earth will be longer.

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?
The length traveled is shorter than the stationary frame, if it were not the same then the length would increase along with the time and the Lorentz factor would not correct it. 

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? 1000m = l_pγ -> 1000/γ = l_p l_p = 769m