Number of light units in a16 cm² box. (+/-2 light units)
| ||||||||
Distance of card from light source (+/- 1 cm)
|
Trial #1
|
Trial #2
|
Trial #3
|
Trial #4
|
Trial #5
|
Trial #6
|
Average
|
Statistical
Uncertainty
(+/- Light units)
|
5 cm
|
81
|
100
|
81
|
90
|
74
|
70
|
83
|
15
|
10 cm
|
25
|
39
|
36
|
36
|
36
|
36
|
35
|
7
|
15 cm
|
22
|
25
|
23
|
21
|
22
|
22
|
23
|
2
|
20 cm
|
11
|
16
|
14
|
13
|
13
|
15
|
14
|
2.5
|
25 cm
|
9
|
12
|
10
|
11
|
12
|
11
|
11
|
1.5
|
In order to count the light units within the square, the length and width were counted. Due to the light units partially in the square, the length and width were both estimated to the nearest whole unit. Therefore, the measurement uncertainty is 2 light units. The statistical uncertainty was calculated for each trial by finding the difference in the range data points and dividing that values by two. For example, in the first IV level the following calculations were used: (100-70)/2= 15. There was a 1 cm measurement uncertainty in the distance because the smallest unit marked on the ruler was one centimeter. Additionally, there is human error in marking out the distances and placing the card precisely on the marked distances each time.
 |
| Graph: Raw data |
 |
| Graph: Raw Data With the line of best fit put onto the graph, it is shown that the data is not linearly related (the line does not fit through the error bars) The Data appears to be an inverse relationship. |
I tried to linearize this function (in many different relationships) and the best fit seemed to be inverse. However, it still did not quite fit. I think I need to reevaluate my uncertainty.
No comments:
Post a Comment