Today we measured the thickness of a human hair by diffraction of light.
For a double slit diffraction of a light beam, light and dark interference spots are created on a surface some distance L from the slots.
If the slots are separated by a distance d, the distance from the center bright spot and bright spot #m is y = λL*m/d
We treat the human hair as the separation between openings and if we shine a laser at the hair, we get a corresponding diffraction pattern on a distant surface.
We placed a whiteboard 1 meter away from the hair, to simplify calculations. our λ was 680nm for a red laser. We measured the distance between the center and the 4th bright spot and got y=4.8 cm
Rearranging and plugging in we got d = 680e-9*1m*4/.048 = 5.67e-5m = 56.7 μm
My hair is on the very lower border of human hair thickness according to google. Makes you think I have alopecia or something haha
Monday, October 10, 2011
Sunday, October 9, 2011
Lenses Lab
In this lab, we looked at the behavior of light through a magnifying glass which is a double convex lens.
We used a light box which illuminated a pattern on a piece of paper, and placed the magnifying glass at specific distances away from the light box, and had a blank sheet on which to project the image.
Before we could do anything, however, we needed to measure the focal length of our lens. To do this, we went outside and focused the light from the sun onto the ground until the light entering the lens was focused at a single point. The focal length was simply the height above the ground.
The focal length of our magnifying glass was found to be f = 14 cm +/- 0.5 cm
After that, we placed the light box at various distances (object distance) from the lens, and found the image distance (where the projection on the sheet was most clear) and measured the image height on the sheet. The object height is constant and is simply the height of the printed pattern on the paper on the light source.
We then tried to get an image when the light box was only 0.5*f away from the lens. We were unable to get a image to show up on the sheet at any distance. However, we are able to see a very clear virtual image by looking through the lens from any point.
We used a light box which illuminated a pattern on a piece of paper, and placed the magnifying glass at specific distances away from the light box, and had a blank sheet on which to project the image.
Before we could do anything, however, we needed to measure the focal length of our lens. To do this, we went outside and focused the light from the sun onto the ground until the light entering the lens was focused at a single point. The focal length was simply the height above the ground.
The focal length of our magnifying glass was found to be f = 14 cm +/- 0.5 cm
After that, we placed the light box at various distances (object distance) from the lens, and found the image distance (where the projection on the sheet was most clear) and measured the image height on the sheet. The object height is constant and is simply the height of the printed pattern on the paper on the light source.
| Object distance (cm) | Image Distance (cm) | Image Height (cm) | Object height (cm) | Magnification |
|---|---|---|---|---|
| 5*f = 70 | 22.5 | 1 | 3 cm | 1/3 |
| 4*f = 56 | 24 | 1.5 | 1/2 | |
| 3*f = 42 | 28 | 2 | 2/3 | |
| 2*f = 28 | 43 | 4.8 | 1.6 | |
| 1.5*f = 21 | 85 | 12.5 | 4.2 |
We then tried to get an image when the light box was only 0.5*f away from the lens. We were unable to get a image to show up on the sheet at any distance. However, we are able to see a very clear virtual image by looking through the lens from any point.
Concave and convex mirrors
In this lab, we looked at reflected images in concave and convex mirrors.
In a convex mirror, (curved outward) objects appear smaller than normal and warped. This stays the same regardless of where the object is in relation to the mirror.
In a concave mirror, however, interesting things happen. Close to the mirror, objects look much bigger than normal. This happens at distances closer than the focal length of the mirror. When the object is moved farther away, the image inverts and appears to shrink.
After observing the concave and convex mirrors, we constructed ray diagrams and calculated magnification.
From the ray diagrams, we measured the object and the image height in order to calculate magnification, which is M = obj height / img height.
For the convex mirror, the theoretical magnification was 1.25/.25 = 5
For the concave mirror, it was negative 5 because the image is upside-down. (the rays don't intersect properly because the focal point isn't placed quite right)
In a convex mirror, (curved outward) objects appear smaller than normal and warped. This stays the same regardless of where the object is in relation to the mirror.
In a concave mirror, however, interesting things happen. Close to the mirror, objects look much bigger than normal. This happens at distances closer than the focal length of the mirror. When the object is moved farther away, the image inverts and appears to shrink.
After observing the concave and convex mirrors, we constructed ray diagrams and calculated magnification.
From the ray diagrams, we measured the object and the image height in order to calculate magnification, which is M = obj height / img height.
For the convex mirror, the theoretical magnification was 1.25/.25 = 5
For the concave mirror, it was negative 5 because the image is upside-down. (the rays don't intersect properly because the focal point isn't placed quite right)
Wednesday, September 28, 2011
Reflection and Refraction
In this experiment, we begin looking at light waves and how they refract. We had a light box with a narrow slot to let a thin ray of light out, a protractor, and a semicircular clear plastic disk.
For the first set of data, we had the light entering the flat side of the disk, and exiting the curved side. We recorded the angles of incident and refraction as theta 1 and 2 respectively.
When we plotted the SINE of the angles against each other, we got a nearly perfectly linear trendline.
For our second data set, the incident ray entered on the curved side and exited on the flat side of the half-disk. At 45 degrees for the incident angle, no light exited the plastic disk.
When the sin of these two angles were plotted against each other, the graph was exactly the same, except the axes were reversed.
The slope of theses lines must represent the ratio for the index of refraction of the materials. Since the index n for air is 1 we could easily figure out the refraction index for the plastic.
For the first set of data, we had the light entering the flat side of the disk, and exiting the curved side. We recorded the angles of incident and refraction as theta 1 and 2 respectively.
| Trial | ϴ_1 | ϴ_2 | sin(ϴ_1) | sin(ϴ_2) |
| 1 | 5 | 4 | 0.087156 | 0.069756 |
| 2 | 10 | 7 | 0.173648 | 0.121869 |
| 3 | 15 | 10 | 0.258819 | 0.173648 |
| 4 | 20 | 12.5 | 0.34202 | 0.21644 |
| 5 | 25 | 16 | 0.422618 | 0.275637 |
| 6 | 30 | 19 | 0.5 | 0.325568 |
| 7 | 40 | 26.5 | 0.642788 | 0.446198 |
| 8 | 50 | 31 | 0.766044 | 0.515038 |
| 9 | 60 | 35 | 0.866025 | 0.573576 |
| 10 | 70 | 39 | 0.939693 | 0.62932 |
When we plotted the SINE of the angles against each other, we got a nearly perfectly linear trendline.
For our second data set, the incident ray entered on the curved side and exited on the flat side of the half-disk. At 45 degrees for the incident angle, no light exited the plastic disk.
| Trial | ϴ_1 | ϴ_2 | sin(ϴ_1) | sin(ϴ_2) |
| 1 | 0 | 0 | 0 | 0 |
| 2 | 5 | 7.5 | 0.087156 | 0.130526 |
| 3 | 10 | 16 | 0.173648 | 0.275637 |
| 4 | 15 | 23 | 0.258819 | 0.390731 |
| 5 | 20 | 32 | 0.34202 | 0.529919 |
| 6 | 30 | 48 | 0.5 | 0.743145 |
| 7 | 35 | 63 | 0.573576 | 0.891007 |
| 8 | 40 | 75 | 0.642788 | 0.965926 |
| 9 | 45 | 75 | 0.707107 | 0.965926 |
When the sin of these two angles were plotted against each other, the graph was exactly the same, except the axes were reversed.
The slope of theses lines must represent the ratio for the index of refraction of the materials. Since the index n for air is 1 we could easily figure out the refraction index for the plastic.
Standing Waves on a string
For this lab, our goal was to analyze mechanical waves on a string, specifically standing waves. We were given a frequency generator, a long string, a ruler, some weights, and a mechanical oscillator. This is a picture of our setup. Not visible is the hanging mass, which is over the opposite end of the table.
When a vibrating string produces standing waves, the number of nodes and antinodes that appear depend on the tension in the string, the length of the string, and the frequency of oscillations. By counting the n number of loops, we can easily come up with wavelength because wavelength is simply 2L/n
The velocity of a wave on a string is denoted by v = √(T/µ) where T is the tension in the string and µ is the mass per unit length of the string.
For our first set of data, the oscillating length of string was 133 cm with a hanging mass of 200g. Our velocity was therefore √(200g/1.38g/m) = 12 m/s
For data set 2, we reduced the tension to 100g
When a vibrating string produces standing waves, the number of nodes and antinodes that appear depend on the tension in the string, the length of the string, and the frequency of oscillations. By counting the n number of loops, we can easily come up with wavelength because wavelength is simply 2L/n
The velocity of a wave on a string is denoted by v = √(T/µ) where T is the tension in the string and µ is the mass per unit length of the string.
For our first set of data, the oscillating length of string was 133 cm with a hanging mass of 200g. Our velocity was therefore √(200g/1.38g/m) = 12 m/s
| Freq. | Nodes | Wavelen | |
|---|---|---|---|
| 17Hz | 2 | 2.66 m | |
| 32Hz | 3 | 1.33 m | |
| 46Hz | 4 | 88.7 cm | |
| 63Hz | 5 | 66.5 cm | |
| 76Hz | 6 | 53.2 cm | |
| 109Hz | 8 | 38 cm |
For data set 2, we reduced the tension to 100g
| Freq | Nodes | Wavelen |
|---|---|---|
| 27Hz | 2 | 2.66 m |
| 48Hz | 3 | 1.33 m |
| 71Hz | 4 | 88.7 cm |
| 88Hz | 5 | 66.5 cm |
| 113Hz | 6 | 53.2 cm |
| 134Hz | 7 | 38 cm |
Friday, September 23, 2011
Mechanical waves
For this lab, we were supposed to come up with an experiment to relate period to wavelength.
The materials we had to use were a stopwatch, a long spring, and a meter stick.
We decided that we would hold the spring at a specific length, and then start a standing wave on it, and measure the time for 10 cycles and divide by 10 to get the period of oscillations.
By changing the distance between the 2 people holding it, and the number of nodes, we could determine the relationship between period and wavelength.
We got the following results:
8 meters 1.04 sec
4 meters .587 sec
2.66 met. .350 sec
2 meters .262 sec
When we plotted this on a graph, the curve was very linear which makes sense since
T=1/f and f=v/λ so T = λ/v
Here is a video of our method.
The materials we had to use were a stopwatch, a long spring, and a meter stick.
We decided that we would hold the spring at a specific length, and then start a standing wave on it, and measure the time for 10 cycles and divide by 10 to get the period of oscillations.
By changing the distance between the 2 people holding it, and the number of nodes, we could determine the relationship between period and wavelength.
We got the following results:
8 meters 1.04 sec
4 meters .587 sec
2.66 met. .350 sec
2 meters .262 sec
When we plotted this on a graph, the curve was very linear which makes sense since
T=1/f and f=v/λ so T = λ/v
Here is a video of our method.
Wednesday, September 7, 2011
Fluid dynamics lab
Our second lab we did was investigating fluid dynamics using Bernoulli's Equation. Our job was to calculate the amount of time it would take to empty a specific volume of water from a larger bucket with the setup as shown:

Using a stopwatch, we measured how long it took to fill up a 200 mL beaker, and repeated the measurement to calculate error. Numbers are in seconds
1. 3.82
2: 4.75
3: 4.35
4: 4.30
5: 4.38
6: 4.52
Using error calculations, we came up with a average time of 4.35 +/- .31
The height of water in the bucket was 144.5 mm +/- 0.5mm
The area of the drain hole was 3.85e-5 m^2 +/-0.025e-5
Calculating the theoretical time with uncertainty yielded a time of 3.48s +/- .39
Comparing the theoretical with the actual value, we have a difference of 25% which is acceptable considering the amount of error introduced from drain hole inaccuracies, and delay in operating the stopwatch.
Calculating backwards, the actual drain hole diameter is 14% smaller than the measured 7mm hole at 5.99mm
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