#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:

Color	Distance 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.

#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.