Color and Spectra

Objective
The purpose of this experiment was to determine the relationship function between theoretical and measured wavelength and use this function to the different between thewavelength of a hydrogen gas spectra

Experiment
             white light source

Spectra observed through a diffraction grating

measurements and calculation of wavelengths

derivation of λ = Dd/√(L²+D²)

a calibration function was determined.

spectra of a hygrogen gas

calculating experimental wavelengths

calibrated wavelengths vs. theoretical wavelengths

Experimental calibrated wavelength (nm)	Theoretical wavelength (nm)
423±25	410
490±14	486
680±27	656

Conclusion

 The expeiment was succeed. All the results are within the uncertainty.

Measuring a human hair

Purpose

 

The purpose of this experiment was to measure the human's hair. We using Young's double slits experiment.

Experiment

    Setting up the equipment:
   


   A laser which produces light rays of around 650 nm wavelength was used in this experiment, and the interference image was projected on a white board.

 

   The length of a slit on the edge was recorded.

   Measure the thickness of a hair with a micrometer

 

Data and Analysis

Wavelength,λ (nm)	633
Distance,L (cm)	150 ± 1
Distance between 2 fringes,y (mm)	11.906 ± 0.01
Experimental diameter of the hair (μm)	79.7 ± 0.65
Actual Diameter of the hair (μm)	65 ± 10
% error (%)	22.6

Conclusion

       This experiment was successfully finished. The result has an error of 22.6% which is good.We learned how to use  Young's double slits to measure a object thickness.

  

 

Determine the Plank's Constant

Objective
In this experiment, we will view the spectrum of colors found in white light and measure the wavelength of several different colors

 

Experiment  

 

 Set up the LED and rulers

 

Measure the length of color from each LEDs

 

Use the mutimeter to measure the voltage across the LEDs and multiply by (e-) to get the energy

 

Datas and results

 

 

 

 

Conclusion

 

In this experiment, we measured the wavelengths of the diffraction lights. And we graph the E vs c/lambda. The slope of the graph should be the Plank's constant and our error is about 15% which is acceptable.

 

Classical Harmonic Oscillator

Objective

The purpose of this experiment was to study quantum mechanics in classical harmonic oscillator with a simulation.

Experiment

Potential Energy Diagrams

 

Potential energy diagram
1&2.The range of motion is from -5cm to 5cm because the energy the particle is bigger than well in this region.
3. The more time the particle spends in one region, the more likely it is to be detected in that region. The particle spends more time to the left of zero because its kinetic energy (and hence its speed) is much smaller in that region. Therefore, the particle is much more likely to be detected to the left of zero.
4. The turning points move outward from the origin by a factor of the square root of two because 1/2 kx^2 = U
5. The shape of the kinetic energy is a parabola, with the opending down.
6. At the turning point. Because the kinetic energy of the particle at the turning points are zero, it is easier to be detected.
Potential Well



1.
E = n² h² / 8 m L²
= (1)² (6.626 x 10^-34 J s)² / 8 (1.673 x 10^-27 kg) (10 x 10^-15 m)²E = 3.3 x 10^-13 J
= 2.1 MeV

No, it is different from finite well.

2.
E = n² h² / 8 m L²
= 4 (2.1 MeV)
= 8.4 MeV

No, it is not the energy of the first excited state from finite well.

3. The wavelength of the wavefunction is larger in the finite well than infinite well, because the wavefunction can penetrate into the "forbidden" regions.

 

4. The energy will decrease. The wavelength of the wavefunction is larger in the finite well than infinite well, so on the same state, the energy of finite well is smaller than infinite well. It shows that as the depth of the well decreases, the energy decreases.

5. The penatrate depth will decrease. As the mass of the particle increases, the chance to penentrate through the forbidden region decreases. Finally, when it is up to a large enough mass, it is consistent with the classical harmonic oscillator in macro scale.

Relativity

Objective

The purpose of this experiment was to study the changing of time and  length under the special theory of relativity.
 
Time Delay

Question 1: Distance traveled by the light pulse
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?


      1. The distance traveled by the light pulse on the moving light clock is longer than the distance traveled by the light pulse on the stationary light clock.

Question 2:

       Time interval required for light pulse travel, as measured on the earth
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?

    

2. The time interval for the light pulse to travel to the top mirror and back on the moving light clock is longer than the interval on the stationary light clock because the distance is longer in the moving clock and speed of light is constant.

 

    

Question 3: Time interval required for light pulse travel, as measured on the light clock
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? Run the simulation, which this time displays timers at rest on both the light clock and the earth.

 

 3. 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 minimum time interval between events is referred to as the proper time.



    

Question 4: The effect of velocity on time dilation
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?
 

 4. As the speed of the light clock is reduced, the difference between the distance traveled by the light pulse and the distance between the mirrors decreases. As this distance difference decreases, the time difference also decreases. Comparing the images below and above, the time difference decreases as the gamma value decreases. When gamma value decreases, the velocity of the traveling light clock decreases, and the time difference 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 (γ) of 1.2.
 

Δt = γΔt_{proper =} 1.2(6.67 µs) = 8.004 µs predicted

actual Δt= 8.00µs very close to the predicted, 8.004s.

     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?  

Δt = γΔt_proper  

Δt = 7.45 µs; Δt_proper  = 6.67 µs

γ =1.12

 

 

Length Contration
 
Question 1: Round-trip time interval, as measured on the light clock
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?

      1. The measurement of this round-trip time interval depend on whether the light clock is moving or stationary relative to the earth.

    

Question 2: Round-trip time interval, as measured on the earth
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?

 

2. The round-trip time interval for the light pulse as measured on the earth be longer than the time interval measured on the light clock.

 

    
Question 3: Why does the moving light clock shrink?
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?
 
3. In order for the time intervals to obey this law, the length of the moving light clock had to be made smaller.
 
 
Question 4: The length contraction formula
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?
     
4. Lp = 1000m, γ =1.3. L=?
         L= Lp/γ = 769m

proved!

 

Conclusion
 

This is a very useful lab and the website helps us to learn the special theory of relativity.

Polarization of Light

Objective
The purpose of this experiment is to observe the light intensity through polarizing filters and measure the changing intensity as a function of angle between the axes. 

Pre- lab question
1. The light intensity is decreasing and at 90 degree it will dark, which means the light can not though the filter.
2. The light intensity is decreasing until it becomes dark and then starts increasing to become very bright.

Experiment

polarizing filters

 

Two polarizing filters. at 90 degree got dark

set up the three polarizing filters 

From this picture, we can see the light pass though the first filter, we get stright light,  the light get darker and darker, but the 45 degree is the maximum

 Data



graph 1




Three polarizers:








The center of axis of the first polarizer is 0 and 90 for the third polarizers. If the second polarizer is on its 0 or 90, the light intensity will be the minimum. If the second polarizer is at 45 orientation, then the intensity observed will be the maximum.

I0*cos(theta)^2*cos(pi/2- theta)^2



I0*cos(theta)^2*sin(theta)^2

1/4 I0*sin(2*theta)^2

when theta= 45, it reaches the maximum value.

 

 

Conclusion:
In part one, when the first polarizer is set to 0 degree and the second polarizer is perpendicular to the first one, the intensity of light will be minimum if the third polarizer is perpendicular to the second one. But if the second one moves to 45 degree, the intensity of light will be maximum. Light intensity depends on the angle of polarization. The light intensity is maximum or minimum as long as polarization is parallel or perpendicular. 

In part two, based on the same reason, first one at 0, second and third are 45 degree. that is the maximum.

CD Diffraction

Objective:
Find the Diffraction of a CD.

  Shoot the laser through the detraction gridding and measure the distance between dots to find the lambda 

 

Measuring the delta x

 Data:

 

 

 

 

Conclusion:

We verified the equation D*sin(theta) = n*lambda. To be accurate, we first us tangent inverse to find the angle and plug into sine function in stead of using small angle approximation sine = tangent.

Our error is about 8% which is not bad and within uncertainty.