We could observe that the angle of incidence of the ray was initially at zero. This meant that the angle of refraction was zero as well, causing the ray to travel back and forth in the same line.
We predicted that the light ray bends at the curved edge of the prism, because the wavelengths are sufficiently small with respect to the dimensions of the prism.
In this experiment light travels from a high density material (transparent prism) to a lower density material (air).
We proceeded to set up the equipment, and found out that the predictions were accurate.
We started to gather data and shift the angle in intervals of 7 and 6 degrees.
Trial | θ₁ (deg) | θ₂ (deg) | sin(θ₁) | sin(θ₂) |
1 | 7 | 7 | 0.121869343 | 0.121869343 |
2 | 13 | 8.5 | 0.224951054 | 0.147809411 |
3 | 19 | 13 | 0.325568154 | 0.224951054 |
4 | 25.5 | 17 | 0.430511097 | 0.292371705 |
5 | 31 | 21 | 0.515038075 | 0.35836795 |
6 | 36 | 22.5 | 0.587785252 | 0.382683432 |
7 | 42 | 26.5 | 0.669130606 | 0.446197813 |
8 | 51 | 31.5 | 0.777145961 | 0.522498565 |
9 | 58 | 33 | 0.848048096 | 0.544639035 |
10 | 66 | 35.5 | 0.913545458 | 0.580702956 |
For the second graph, the slope came out to be 1.6258 which is the ratio of the index of refraction.
For the second case, we inverted the prism so that the rounded part would face the ray of light.
We assumed that if a light ray hit the curved surface first at the initial position showed above, the ray would not be distorted. It would travel back and forth along the path of the ray.
When the ray strikes the flat surface, it will refract and it will bend the ray by an amount of degrees.
The ray is still traveling from a high density to a lower density medium.
This time we made intervals of 6 and 7 degrees in between trials.
Trial | θ₁ (deg) | θ₂ (deg) | sin(θ₁) | sin(θ₂) |
1 | 0 | 0 | 0 | 0 |
2 | 8 | 2 | 0.139173101 | 0.034899497 |
3 | 14 | 4 | 0.241921896 | 0.069756474 |
4 | 21 | 9 | 0.35836795 | 0.156434465 |
5 | 27 | 14 | 0.4539905 | 0.241921896 |
6 | 34 | 20.5 | 0.559192903 | 0.350207381 |
7 | 41 | 40 | 0.656059029 | 0.64278761 |
8 | 48 | 84.5 | 0.743144825 | 0.995396198 |
9 | 55 | 72.5 | 0.819152044 | 0.953716951 |
10 | 62 | 55.5 | 0.882947593 | 0.824126189 |
11 | 70 | 41 | 0.939692621 | 0.656059029 |
When graphing sin(θ₁) vs sin(θ₂), a portion of the graph had a linear trend. The slope of the linear region came up to be 1.485 which was different to the slope obtained from case 1 (1.6258).
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