ActivPhysics: Relativity

Relativity of Time
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?

A: It's longer than the distance on the stationary 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?

A: The time is longer than it will take compare to the one on the stationary station.When the light clock is moving, the light pulse must travel along the hypotenuse of a right triangle whose legs are formed by the distance that the light clock moves and the distance between the light clock's mirrors. 

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?

A:If I am the one on the light clock, the time i measured will be the same as the stationary light clock. Since the light pulse always travels the same distance, it always takes the same amount of 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?

A: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.

A: The time the outside observer observed appeared to decreased as the velocity of the light clock moving increase.

Question 5: The time dilation formula

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.

Set γ = 1.2 and run the simulation to check your prediction.

A: It's 8.00microseconds! The proper time interval does not depend on the speed of the light clock. The time interval measured by an earth-bound observer is the product of the proper time interval and the Lorentz factor.

Question 6: The time dilation formula, one more time
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?

A: The Lorentz factor is the ratio of the observer's time interval measurement to the proper time interval,  (7.45 µs) / (6.67 µs) = 1.12.

Relativity of Length
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?

A: The measurement of this round-trip time interval depends 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?

A:The round-trip time interval for the light pulse as measured on the earth will be shorter 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?


A: NO

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?

A: L = 1000/1.3m = 769m

#12: CD Diffraction

The purpose of this lab was to determine the distance between data pits on a CD. This was done using the properties of diffraction, or more specifically, diffraction grating. A 630-nm laser source was placed 0.51±0.005m away from a CD so that its laser had a near-normal incidence on the CD. The reflected diffraction pattern appeared on a whiteboard that was placed next to the source. Since the laser had near-normal incidence, even the zero-order maxima appeared on the edge of the whiteboard.
 First set up the experiment with a lhelium neon laser perpendicularly to the surface of the disk. As the laser hit the surface of the CD, the laser beam diffract and split into multiple beams of light. Both the CD and the laser should be adjusted until the the zero order maximum would shown on the board. Then record the the perpendicular distance between the board and the disc and the distance between the maximums. 

Data and Analysis
tan(θ) = x/Ld=mλ/sin(θ)

Conclusion

This experiment let us have a better understanding of laser diffraction and how it can be used to measure defects in CD spacing. However, there are some errors for the data we collected in the experiment. Since we hold the screen between the laser and CD by hand, and we do not hold the screen to be perpendicular to the laser and CD, which makes the measured distance between the CD and screen L and calculated distance x not accurate. With more precise instrumentation would could significantly decrease our percent uncertainty and difference to more accurately the CD's spacing.

#6: Measuring the Length of a Pipe with Standing Waves

Objective: Use the frequency of sound emitted by a swinging pipe to determine it's length using sound wave and standing wave equations.


Procedure:

1) Swing the pipe in a circle at a constant angular velocity in order to create a standing wave in the pipe. The frequency will be recorded using a digital microphone.

2) Accelerate the pipe until it emits the next harmonic frequency. This frequency will be recorded using the digital microphone.

3) Determine the length of the pipe using these two frequencies assuming that the speed of sound remains constant.


Calculations:
Omega equals to 3859 rad/s for the first case, so f-1 roughly equals to 614 hz;  for the second case, omega = 5068 rad/s; thus, f-2 equals 806 hz.

n = harmonic number of lower frequency, L = wavelength

(2n + 1)(L1)/4 = (2n + 3)(L2)/4

L1 = v/f1   L2 = v/f2

(2n + 1)(f2) = (2n + 3)(f1)

n = 2.7 ~ 3

Length of pipe = (2n + 1)v/(4f1) = 0.978 (v = speed of sound assumed to be 343 m/s)

#11: Measuring a human hair

Objective:
Measure a human hair by using the principle of single slit diffraction producing predictable minima.


Materials:
A human hair, a hole punched card, and a laser pointer.

Procedures:

1. Mount a strand of hair to an index card with a hole in it

2. Shine a laser passing through the hair

3. Applying the equation d = (lamda*L)/y

4. Measure the parameter y and L, with the known parameter lamda of laser = 632.8 nm = 632.8*10^-9 m

5. Find d - the thickness of hair

Data obtained:

The data obtained by the laser experiment is as follow:

Wavelength (nm)	633
Distance to the board, L (cm)	100 ± 0.5
Distance of fringes, y (cm)	1.05 ± 0.02

d =  632.8*10^-9/1*10^-2 = 6.33*10^-5 m = 0.0633 mm

Conclusion:

The average hair thickness of online sources(expected value): 0.09 mm - 0.25 mm

Out result does not fall within this range, but it is in the same order of magnitude.

Error(lower): abs(0.0633-0.09)/0.09 = 30 % 

Error(upper): abs(0.0633-0.25)/0.25 = 75 %

* The use of micrometer to measure the thickness of human hair is not an accurate method, as least compared with out main experiment, because of the limitation on the magnification of the equipment compared with the actual order of the thickness.

#8 Lenses

Objectives
The objective of this experiment is to determine a relationship between the object distance d0, the image distance di, and the image height hi. This can be accomplished by increasing the focal length by an integer value each trial. We began this experiment by measuring the focal distance of the lense that we were using. To do this we used the sum as our source and moved the lens until the light was focused at a single point. We then measured the distance from this point to the lens. We determined that the focal distance for our particular lens to be about 20.0 centimeters. We then shined an image through our lens and measured the object distance, image distance, object height, image height, and we calculated the magnification. We also stated whether or not the image was inverted.

Materials:

Socket lamp with V-shaped filament

Large converging lens

Large split lens or masking tape

Lens holder for large lens

cardboard

Meter stick

Procedures:

1. (a) Put the object in front of the convex mirror and observe the size and the orientation of the image

    (b) Observe the location of the image

    (c) Observe the change in image when moving the object closer or further from the mirror

2. Repeat #1 for concave mirror

3. Complete the ray diagram worksheet for the both convex and concave mirror; compare the result with the observation.

Data and Analysis

Focal length (f) = 15.5 ± 0. 5 cm

Table 1: Recorded data of the object and image distances and heights

Object distance as a multiple of f(cm)	Object distance(cm)	Image distance(cm)	Object height(cm)	Image height(cm)	M	Type of image
5f	77.5 ± 0.5	28.70 ± 0.5	8.50 ± 0.2	3. 0 ± 0.5	0.35	Diminished, Inverted, real
4f	62.0.0 ± 1.00	28.5 ± 0.5		4.20 ± 0.2	0.494	Diminished, Inverted, real
3f	46.5 ± 0. 5	34 ± 0.5		6.40 ± 0.5	0.75	Diminished, Inverted, real
2f	31.0 ± 0.20	51.70 ± 0.2		14.50 ± 1	1.71	Same size, Inverted, real
1.5f	23.0 ± 0.5	96 ± 2		38.2 ± 2	4.49	Magnified, Inverted, real

Observations:

- For convex mirror, the image formed is smaller than the object, and it is upright. The image seems to locate    deep inside the mirror. As I move the object closer to the mirror, the image is getting larger and closer to the object's size, and it is getting smaller when I move the object away.

- For concave mirror, the image formed is larger than the object, and it is inverted. The image seems to locate not as deep as the convex mirror. As I move the object closer to the mirror, the image is getting smaller and closer to the object's size, and it is getting larger when I move the object away. The orientation would change as I move the object back and forth.

#7 Concave and Convex Mirrors

There are three types of mirrors, plane, concave and convex.

 convex

 concave

plane

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