Monday, June 11, 2012

#13 Relativity

Part 1: Time
Q1: The light traveling on the moving clock passes longer distance than the stationary clock.

Q2: The time required for the moving clock traveling back and forth is longer than the stationary clock.

Q3: On the moving clock, the distance and time is the same as the stationary.

Q4: If the velocity of the moving clock decrease, the difference in time also decreases.
Q5: t = 1.2x(2000/(3x10^8)) = 8x10^(-6), the simulation caters to the calculations.
Q6: 3 x 10^8 x 7.45 x 10^(-6) / 2000 = 1.12

Part 2: Length
Q1: Measuring in the frame, the time is the same for moving and stationary.
Q2: The time measured on earth is longer than the measurement in the light clock.

Q3: They will not be equal.
Q4:  1000 / 1.3 =769 m












#15 Laser

Q1: 3+183+20=206 for this specific moment. Other moments also shows conservation.

Q2: There is no preferred direction for which the photon goes.

Q3:  Although most of them emitted at the beginning, there can be a few atoms remaining in its exited states for a long time,so the time is not constant.

Q4: When the photon interact with the excited atom, the atom emits another photon that is in the same direction and same phase as the incoming photon.

Q5: 70 might be a good guess because it varies from 8:12 to 12:8 nearly all the time.

Q6: The odd direction photons are in spontaneous emission originally, but these photons sometimes interact with another exited photon thus causing a stimulated emission of photons.
  



Thursday, May 31, 2012

#14 Spectra


Objective: 
The purpose of this experiment is to measure the wavelength of a few visible light and form a linear transformation of the derived wavelength, and use this to measure the spectra of hydrogen atom and an unknown to find out what the unknown is.




Procedures:

1.      Set up as instructed in the lab manual, put the light source as the white light.
2.      Record the position for each light seen through the diffraction grating. 
3.      Change the light source to hydrogen atoms and unknowns and perform same procedure.


Results:

       The acceptable visible light range is from 390 nm to 750 nm.

The wavelength is derived using formula:




L is 100 cm, and d is 1/500 mm.
For white light:

Color Distance D (cm) λ(nm)
Purple Begins 17.5 345
Purple -> Blue 22 430
Blue -> Green 24.5 478
Green -> Yellow 26 503
Yellow -> Red 28 539
Red 35 661
Red ends 40 743

           Using 345 and 743 to do the linear transformation, the equation is λ = 0.905λ'+78 in nm.


Then for the hydrogen atom:
ColorDistance D (cm) λ(nm)
red 33.5 653
Yellow 28 566
Green 24 500
Purple 21 450
            
λ is derived from the early equation and linear transformation.

For the unknown, we derived following data:
Trial 1:
Color Distance D (cm) λ(nm)
Yellow 29 582
Green 27 550
Blue 21 450
Trial 2:
Color Distance D (cm) λ(nm)
Yellow 29.5 590
Green 28 566
Blue 21.5 458
            We suspect the unknown to be Mercury (Hg). The spectra for Mercury is:
          The spectra for Hg shows peak at 435, 545, and 580 nm, which is very close to the calculated data.

Discussion:
          The measurement is very inaccurate for the precise distance from diffraction grating to the light source is hard to measure, so does the displacement. The derived spectra for white light gives a fairly precise linear transformation to the system. The unknown is Hg as what professor told us, but the wacelength for blue light is comparably different from the theoretical one.

Wednesday, May 30, 2012

#12 light and matter waves


Objective: 
          The purpose of this experiment is by using Vpython to visualize the wave in the space.
          By creating a wave source at 0,0 producing a wavelength of 2mm, it yields as following:
     Contour as shown:

          Varing the wavelength to 4mm and 8mm, yields:
8mm 
4mm
          They show nearly the same pattern, and the peak is at the wave source.

          For double slits, the separation is 1.2cm, wave sources are set at (0,0.6) and (0.-0.6).
          Wavelength in 2mm, 4mm, and 8mm yields interference pattern as following:
2mm

4mm

8mm
          Thus again they perform a likely pattern. The intensity graph shown is:
          Shows it reaches peak near the central point, that is what shows in the graph.

          






#11 CD Diffraction



Objective: 
The purpose of this experiment is to measure the length of the gap between the grooves on the CD and DVD.

Procedures:

1.      Arrange the equipment as guided, shoot the laser to the CD/DVD.
2.      Record distance between two diffracted points on the white board behind the laser, make sure the laser is in the middle.
3.      Record wavelength of the laser and the distance from the board to CD/DVD.


Results:

       The wavelength of the laser is from 630 to 680 nm. Formula is given as:

       d*sin(theta)=d*y/L=n*lambda
       For CD:
        The distance from CD to screen is 47.5cm, distance between two point on the board is 49cm.Thus the angle is 26.32 degree.
    L = 47.5cm, theta = 26.32, y = 24.5cm, lambda from 630 to 680 nm.
    Derived value for d is from 1432 to 1545 nm, the proper distance from internet is 1500, which is in the range.
    For DVD:
    L = 20cm
    y = 19 cm
    Angle theta = 43.5
    Lambda unchanged
    d is from 915 to 988 nm. Proper distance is 740, which is not in the range.

       

Discussion:
    The measurement conducts well for CD. The range is mainly due to the uncertainty of the laser wavelength, but the theoretical value falls in the range. The experiment conducts not so well for the DVD, which does not fall in the range. Maybe DVD has different groove structure or different arrangements, which may cause error in this method of measurement.

Thursday, April 12, 2012

#9 Microwave oven


Objective: 
By the sacrifice of brave marshmallows we shall be able to determine the frequency of the microwave by assuming a standing wave is producing in the oven. Furthermore, a rage of microwave shall be deduced in the dimensions of microwave oven. By heating up water we can acquire the power and then decided how many photons per seconds are oscillating in the microwave and what pressure do these photons exerts.

Data recorded:
      λ= 12 cm, dimension of the oven is 23 x 35 x 35 cm3, mass of water is 100 g, heating time is 30 s, Ti is 20 , Tf is 57 .


Calculation:

       The wave is electromagnetic wave, thus v = c.

       c =  λf , thus f = 3 x 108 / 0.12 = 2.5 x 109 Hz
       Cwater = 4.186 J/(g)
Thus the energy is E = Cwater x mwater x (Tf - Ti) = 15488.2 J
P = E / t = 516.3 W
Energy of photon is Ep = h f = 6.626 x 10-34 x 2.5 x 109 = 1.65 x 10-24 J
Thus, total oscillating photon numbers are P / Ep = 3.13 x 1026
Assume power exerts evenly on the inner surface, then
Area = 2 x (0.23 x 0.35 + 0.23 x 0.35 + 0.35 x 0.35) = 0.567 m2
Intensity is I = P / A = 516.3 / 0.567 = 910.6 W/m2
I = Ppressure x c, therefore Ppressure = 910.6 / 3 x 108 = 3.04 x 106 Pa

#8 Concave and convex mirrors


Objective: 
Give a qualitative observation of the image formed by both convex and concave mirrors.

Procedures:

1.      Describe the image formed by placing an object in front of a convex mirror.
2.      Move the object and describe the change of image.
3.      Repeat the process for concave mirror.

Results:

       Convex mirror:

       The image appears to be smaller and away from the center compared with object, it is not inverted. When moving towards the mirror the image grows larger and more like the original size.




 

       Diagram as shown:


do = 15 cm, ho = 10 cm, d­i = -7 cm, h­i = 5 cm.
       Agrees with observation.

       Concave mirror:
       The image appears to be much larger than the object and more close to the center, but still not inverted. When moving towards the mirror the image becomes smaller while moving away from the mirror the image gain larger and larger and disappear then become inverted.

       Diagram as shown:

      do = 20 cm, ho = 8 cm, di = 12 cm, hi = -5 cm
       The diagram shows the situation when the object is further away from the mirror, which caters to the latter description.

Discussion:
       The theoretical diagram agrees with the observation. Experiment conduct successfully.