First we found the focal point of our given lens. To do so we tried to focus the sunlight on a point and then we measured the distance from the center of the lens to that point. Our value was averaged to f = 25 ± 1 cm
the error is so high since at that time we had to hold the lens at a very big angle relative to the normal of the ground which made measuring values harder.
Then we used the same lens in the following setup to find the desired distances:
| Object Distance d0, cm | Image Distance di, cm | Object Height h0, cm | Image Height hi, cm | M | type of Image |
| 74.85±0.1 | 23.5±0.1 | 3±0.1 | 1.3±0.1 | 0.43±0.05 | inverted inverse |
| 59.88±0.1 | 24.5±0.1 | 3±0.1 | 1.75±0.1 | 1.75±0.05 | inverted inverse |
| 44.91±0.1 | 28.7±0.1 | 3±0.1 | 1.95±0.1 | 1.95±0.05 | inverted inverse |
| 29.94±0.1 | 44.5±0.1 | 3±0.1 | 5.2±0.1 | 5.2±0.05 | inverted inverse |
| 22.46±0.1 | 332.5±0.1 | 3±0.1 | 53.6±0.1 | 53.6±0.05 | inverted inverse |
the graph clearly shows that d_0 and d_i have an inverse relationship of some sort. So if we graph the 1/d_0 vs 1/d_i we obtain the following graph:
this graph fits to the linear equation y = -1.2654x + 0.0617
the absolute value of -1.2654 is 1.2654 which is close to index of refraction of air however the value has a huge error due to big uncertainties in focal point and the fact that the fitted line is only using 5 points which is not clearly enough.
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