We began by first finding out the focal length of the lens. To do this, we needed a light source that would offer infinite object distance, so we took the Sun as our light source.
We measured the focal length of the lens to be 4.8±.03 cm. Once the focal lenght was known, we set up a meter stick that worked as a rail on which the lens would be attached. The light source would make a projection that would go through the lenses, showing the shape of a plastic film in front of the light box.
Having the focal lenght, we started placing the lens in intervals along the meter stick. First we started by placing the lens 4 focal length distances away from the light source.
The distance was measured to be do = 20.0±.3
cm.
We also measured the distance from the lens to the image on the clear surface to be di = 5.4±.2
cm. The height of the filament was recorded to be ho = 9.2±.2 cm, whereas the height of the projected image was measured to be hi = 2.9±.2 cm.
The magnification is equal to:
hi/ ho
= 2.9±.2 / 9.2±.2 = 0.32±0.023
Qualitatively, the projected image appeared inverted. The projection was a real image of the object.
Reversing the lenses did not changed the measurements obtained above.
We then changed the distance so that the lens would be 2 focal length distances away.
This distance now was do = 10.0±.2
cm. with an image distance of di = 8.6±.2
cm.
The heights were measured to be the same ho = 9.2±.2 cm. for the object, and hi = 7.3±.2 cm. for the image, and magnification M = 0.79±0.028.
The displayed image seemed to get bigger now that we were closer to the filament. Again, nothing changed when reversing the lens.
With now a distance of 1.5 focal lengths, the distance changed to be do = 7.5±.2 cm. and di = 10.5±.3 cm.
The heights measurements for the object were ho = 9.2±.2 cm and hi = 13.0±.1 cm. and M=1.41±0.033 with no change in the reversed lens technique.
If the lens were to be covered in half, I predicted that only half the image would be visible. This turned out to be not right. When doing so, the complete image still went through the lens, but the projection was dimmer. This is probable due to the light rays traveling around the lens, but with decreased intensity due to the covered up portion.
dâ‚’ (cm) | d₁ (cm) | hâ‚’ (cm) | h₁ (cm) | M | Type of image |
5f = 25.0±.2 | 6.0±.2 | 9.2±.2 | 2.1±.1 | 0.23±0.012 | Real/Inverted |
4f = 20.0±.2 | 5.4±.2 | 9.2±.2 | 2.9±.2 | 0.32±0.023 | Real/Inverted |
3f = 15.0±.2 | 6.4±.2 | 9.2±.2 | 3.8±.2 | 0.41±0.024 | Real/Inverted |
2f = 10.0±.2 | 8.6±.2 | 9.2±.2 | 7.3±.2 | 0.79±0.028. | Real/Inverted |
1.5f = 7.5±.2 | 10.5±.3 | 9.2±.2 | 13.0±.1 | 1.41±0.033 | Real/Inverted |
dâ‚’ (cm) | d₁ (cm) | hâ‚’ (cm) | h₁ (cm) | M | Type of image | Inverse d₁ (cm^-1) | Neg. Inverse dâ‚’ (cm^-1) |
5f = 25.0±.2 | 6.0±.2 | 9.2±.2 | 2.1±.1 | 0.23±0.012 | Real/Inverted | 0.167±.0056 | -0.0400±.00032 |
4f = 20.0±.2 | 5.4±.2 | 9.2±.2 | 2.9±.2 | 0.32±0.023 | Real/Inverted | 0.185±.0069 | -0.0500±.00050 |
3f = 15.0±.2 | 6.4±.2 | 9.2±.2 | 3.8±.2 | 0.41±0.024 | Real/Inverted | 0.156±.0049 | -0.0667±0.00089 |
2f = 10.0±.2 | 8.6±.2 | 9.2±.2 | 7.3±.2 | 0.79±0.028. | Real/Inverted | 0.116±.0027 | -0.100±.0020 |
1.5f = 7.5±.2 | 10.5±.3 | 9.2±.2 | 13.0±.1 | 1.41±0.033 | Real/Inverted | 0.095±.0027 | -0.133±.0036 |
When analyzing the data in this last graph, I had to take out the point (-0.0400±.00032,0.167±.0056) out of the population so that it would yield a better trendline. I treated that last point as foul data to yield a slope of 1.075±.1454. The y-intercept came out to be 0.2379±0.01353. This suggests the relationship 1/f = 1/s +1/s', and a expression of the form d₁ = 1.075±.1454dâ‚’ + 0.2379±0.01353.
Potential sources of error for this experiment are the amount of uncertainty when trying to get perfect focus in measuring d₁. Determining the focal length can also influence the change for getting large uncertainties, and extra-ordinary data points in the graphs.
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