The independent variable in this lab is the angle of the incident ray, the responding is the angle of the refracted ray. The controlled variable include - the intensity of the ray, the position of the plexiglass, and the apparatus. To control the intensity of the ray, use the same type of material. The Plexiglass should be kept at the center of the graph paper nas the apparatus was kept constant by using them again .
The materials that would be required is - a light source, a plexiglass, a graph paper which tells the angle.
Procedure:
1. Obtain all the materials and set it up as shown in the above diagram.
2. Keep the light source 1cm away from the start of the graph paper vertically so that the angle at which the light is going in is visible.
3. Pass the light and record the incident and the refracted ray.
4. Repeat with at least 5 other angles.
5. Repeat to confirm results.
6. Clean up.
Data Collection and Processing:
Table #1: Raw data includes the angle of incidence and angle of refraction
Calculating the sin of the incident angle and the refracted angle:
sin(10°) = 0.1736481777 = 0.17
Sin (173°) = 0.1218693434 = .12
The error of the sin of the angles is exempt
Table #2: Sin value of the incident angle and the refracted angle
Fig 1: The graph of the sin of the incident and the refracted angle is shown below. The slope in this graph represents the index of the refraction of the plexiglass, since the n value of air is 1.00. The graph below does start from 0, however the y value is 0.02. This is because when the incident angle is 0 degrees, the refracted angle is 179 degrees, instead of 180 degrees.
The graph above shows us the slope to be 0.9629, however to confirm the results the calculations are done below. There are no errors in the above graph, therefore there will be no max slope and min slope calculations.
Calculation for the slope:
m = y2 - y1 / x2 - x1
m = (0.874619707-0.777145961)/(0.866025404-0.766044443)
m = 0.9629
The inverse of the slope would be taken as the above slope calculates refracted over incident. Therefore, the slope and therefore the index of refraction of plexiglass is
1/m = n
1/0.9629 = 1.038529442
N = 1.04
The experimental index of refraction is found to be 1.04, however the theoretical value is 1.51. Therefore, the percent discrepancy must be found.
1.04 - 1.51 = 31.1%
1.51
The percent discrepancy is 31.1%
Conclusion and Evaluation:
This aim of this lab was to find the index of refraction of the plexiglass. This was done by passing a ray of light at the plexiglass, at a certain angle and then recording the angle at which the ray is refracted back into the air. The index of refraction was found to be 0.96 however there was a lot of percent discrepancy, 31.1% found. The index of refraction was higher than the air of refraction of air, which tells us that when the ray was passed it bended towards from the normal, meaning that the plexiglass was a more optical medium.
There were some weaknesses found in the lab. One weakness found was that that it was hard to keep the plexiglass at the same place. This could be solved by putting the plexiglass at a fix distance so that it won’t move. Another weakness was that the plexiglass had too many scratches and it was also dirty. This could have affected the results. This could be solved by either cleaning the lens with the lens cleaner or get a new glass.
