Experiment 12: CD diffraction

When you look the DVD into a micro structure level, you can see there are little curb on the surface of the DVD. By using the diffraction method, we can simply measure the distance between the grooves on the DVD.
The one we have was a DVD, and the method we use is to point the laser at a angle to the DVD. A beam of laser will diffraction on the other side of the DVD where we put the white board to measure the distance between the first order maxima

Even thought its hard to see whats happening in the picture, we can see the light been goes into the DVD and another light beam comes out from the DVD.

Since we know the wavelength of the laser beam, which is around 670nm. Also, the distance between the DVD to the white board is around 0.07m. The distance between the center and the first order maxima is around 2.8cm. By using the diffraction formula, we can calculate the estimate groove distance by using y=m λx/a  where y is the distance between the maxims, x is the distance between the source to the screen, λ is the wave length. In this case, m is 1 since we measured the first maxim. The value we try to find is a which is around 1.675*10^-6 m.  As we compare to the manufacturer's standar value 1600nm, our result is very close to the manufacture's number. The difference between the numbers is due to our measurement. Since we just use a regular ruler to measure the distance between the first maxim, there are a larger uncertainties in it. Also, the distance between the board and the source is also measured with the same method, which gave us a larger uncertainties. With all the uncertainties we have, we can conclude that our measurement is very reliable, and it proved that the DVD has grooves about 1.6*10^-6 m  in length

Relativity - Online animation

Today, we were introduced the basic principle of relativity.
There are two basic concept behind the formula we learned to day, which are speed of light is constant, and the fastest speed you can get is speed of light.

Thus, we learned the time dilation and the length contraction of the object at a high velocity(perhaps half of the speed of light).

The equation of time dilation is 

Thus, we went online to understand more about the time dilation question

Answer q1: The distance travel by the moving light clock is greater than the distance travel by the stationary light clock

AnswerQ2: Since the speed of light is constant and the distance is longer, the time require the light pulse to travel must also be longer than the normal time.

Answer Q3: In you frame of reference, the light pulse does not travel a greater distance, thus the time require for the light to travel is the same time as the light travel in a stationary frame.

AnswerQ4: If the velocity of the light clock decreased, the difference of earth's timers and the light clock's timers also decreases

AnswerQ5: Since we know the loreantz facter (gamma) of 1.2, the time dilation of a moving clock with a lorentz factor of 1.2 will have a 1.2 times of the original time.

AnswerQ6: Since the measure time is 7.45 muS, we math the time in the simulator and found the lorentz factor is around1.12, which is what we expected on the paper


Later, we were introduced the concept of length contraction.

After understood the question, we moved to the online simulator to understand more about the length contraction

AnswerQ1: The round trip time depends on whether the clock is stationary or moving

Answer Q2: The round trip time will measured shorter than the time measured relative to the earth.

AnswerQ3:Yes. If we know the loreanzt factor and the time it required  for a movie clock for reach the distance, we can use the formula to calculate the time it require to reach the same distance relative to the stationary frame.
Answer Q4:If we know the Lorentz factor is 1.3, and we know the length is 1000m at rest, we can use the formula to calculate the contracted length , which is abount 1000/1.3  = 769m

Programming - Exercise 5

Today, we introduced the programming.

For people who have taken 4A with mason, will need to go to the handout #2 of the programming, which is a more advance handout than the other handout.

Since I had 'some' programming experience with mouse trap car, and the space mission to Mars, I had to do the handout #2. Also, I am taking a introduction of C language, which helps me a lot in this package since I learn most of the material in my class already. At the end of the handout, we were suppose to answer the question at the end of exercise 5, which is the create two array, one is sin(2*pi*i/100) the other is cos(s*pi*i/100) where i is a constant from 1 to 100
After created the two arrays, we want to find the dot product of the two arrays, which suppose to be zero since the dot product of a sine and cosine function is zeor

Here is the program I have


from numpy import*
from decimal import*
xx=zeros(100,int)
yy=zeros(100,int)
zz=zeros(100,int)
result=0
for i in range (0,99):
    xx[i]=sin(2*pi*i/100)
    yy[i]=cos(2*pi*i/100)
print xx
print yy
for i in range(0,99):
    zz[i]=xx[i]*yy[i]
    result=result+zz[i]
print result

The first line is to input the mathematical calculation of sin and cos from the library

Later, I declared two arrays of xx and yy to put the calculated varaibe into the two arrays with a loop to input the number into each cell

At the end I declared another array of zz to do the dot product of the two arrays and the result is zero as we expected

10/6 Experiment11: Measuring a human hair

The method behind this experiment is using the diffraction of light to measure the diameter of light

We use the hair to treat it as the difference between the hold to make diffraction of light

The distance from the card to wall is 131.2cm
the measured distance between the two first order minima is 1.7cm
by using the formula y=L入/d
and by finding the value we get 5.17*10^-5 m

If we use a micrometer to measure the diameter, we get 0.037mm which is 3.7*10^-5 m

the numbers are in the same order of magnitude

But the uncertainties for using the hair as a split for the whole is greater because there are more uncertainties   because it is really hard to measure the distance between the two first order minima due to the lack of precision in equipment (ruler). On the other, the micrometer is build to measure precise small length. But it is way cheaper to measure the diameter of hair by using a laser beam and piece of card to measure the diameter of the hair

10/4 Experiment 10: Lenses

First, we went outside under the side to figure out what is the focal length for the magnification glass has, the number we got was 24.5 and 24.3 cm, which the average is 24.4cm.
Later on we use the magnification glass and a light source with a figure in the front to project the image over a certain distance.

The data we obtained

Object distance,cm	image distance,cm	Object height,cm	Image height,cm	M	Type of image
5f=122	30.9	3	0.8	0.267	inverted
4f=97.6	33.3	3	1.05	0.35	inverted
3f=73.2	37.3	3	1.6	0.533	inverted
2f=48.8	48.4	3	3.15	1.05	inverted
1.5f=36.6	70.9	3	5.7	1.9	inverted

If we put the object within the image, we cannot see the image, the image will become infinite

if we graph di vs. do



 the inverse fit on the loger pro does not work perfectly, thus we will consider to plot the inverse di vs. negative inv do

According to formula: 1/do + 1/di = 1/f. As we expected, the 1/y-intercept will give us a value of 25 cm, which is gonna be the focal length.  And the slope of the regression line is suppose to be 1 as we have

 Both the values of focal length are within their uncertainties, the uncertainties from where we obtained under the sun is 24.4 +- 0.2cm . The uncertainties for data uncertainties is 25.1 +- 0.4cm

 The data can be consider to be math to each others within their uncertainties. We can measure the focal length of the magnification glass with two methods, one by using a sun and focus to a point and measure the height, or by using a known object distance and image distance to calculate the focal length. The only problem with the method 2 is when we try to focus the image on a white board, it is really hard to tell if the image is in focus or not when the image distance is far away from the lense due to the decreasing in light intensity.

9/29 Experiment 9: Concave and Convex Mirrors

Today's experiment is pretty simple, just by looking at different type of mirrors and record the results.
There are three types of mirrors, plane, concave and convex.
First, we looked at convex mirror.
The first interesting thing is that the image appear on the mirror is smaller than the actual size of object. Another interesting is that the image is always upright. When we move the object closer to the mirror, the image will become bigger, but still smaller than the object. Vice verse,  when we move the object father away from the mirror, the image seem to become smaller.

Next, we learned how to draw the ray diagram for convex mirror

buy drawing the rays, we can estimate the size of image and calculate out the magnification
for this diagram, the object size is 5.6cm, the image size is 1.9cm   the magnification is 1.9/5.6=0.34


The second part of the experiment is to understand the how concave mirror work
The image is always inverted unless you put the object within the focal point. When you put the image in the focal point, the image will be infinite large, which you cannot see the any image. When you move the object away from the mirror, the image will become smaller. After know the behaviors of concave mirror, we drew the ray diagram for concave mirror

the image is 0.75cm , the object is 3.1cm, thus the magnification is 0.242