Exp 10

Exp 10
Lenses

Purpose:
To observe the behavior of light through converging (convex) lenses and how magnification and height is a function of the distance of the object from the light source; to also observe when an image is inverted and upright in front of converging lens.

Materials:
1. BIG rulers (meter stick)
2. Light source (light box with distinct shape on the opening)
3. Converging (convex) lens and lens holder
4. Flat, not diffused, surface

Procedures:

Since the lenses' focal length was unknown, it had to be predetermined before the lab could continue. This can be done by facing the lens against a light source infinitely far away (to get straight light rays) and measure the distance at which the converging light rays are strongest. 






 The height and horizontal distance from the meter stick was measured, forming a triangle. The hypotenuse would be the distance of focal point (f).

The triangle was a right, 6.0 * 7.0 triangle, with hypotenuse of 9.2 cm, the focal length.





Actual lab and questions:

The lens was put on the lens holder on a meter stick, against a light source. On the other end, a smooth surface was set up to show the shape of the image and measured.

Initially, the lens was put at distance four times focal length (f), which was 36.8 cm.
d_0 = 36.8 cm
The surface was adjusted to get the sharpest image. The distance between surface and light source was 44.2 cm. The distance between surface and the lens,
d_i = 7.5 cm



The height of filament (light box/ source) is
h_0 = 9.0 cm
whereas the height of the image
h_i = 3.1 cm
Magnification, M is h_i / h_0 = 0.33, a third the of original object.
The image is real.


When the lens is reversed, the measurements were still the same.
d_0 = 36.8 cm
d_i = 7.5 cm
h_0 = 9.0 cm
h_i = 3.1 cm
M = 0.33

The lens was moved back, towards the light source, so its distance was now 2f.
d_0 = 18.4 cm
d_i = 30.2 - 18.4 = 11.8 cm
h_0 = 9.0 cm
h_i = 6.0 cm
M = 0.66
The image was still the same as the above data when the lens was reversed.

The lens was moved further back, with distance 1.5 f
d_0 = 13.8 cm
d_i = 31.0 - 13.8 = 17.2 cm
h_0 = 9.0 cm
h_i = 12.0 cm
M = 1.33
Image was the  same reversed.

When half of the lens was covered, the image was still reflected with the same shape, but dimmer.


Lab part 2.
The lens was now set up at different distances.

h_0 = 9.0 cm

5f:
d_0 = 46.0 cm, d_i = 18.6 cm, h_i = 2.2 cm, M = 0.24
Image was inverted.
4f:
d_0 = 36.8 cm, d_i = 12.3 cm, h_i = 3.1 cm, M = 0.33
Image was inverted.
3f;
d_0 = 27.7 cm, d_i = 10.0 cm, h_i = 3.4 cm, M = 0.36
Image was inverted.
2f:
d_0 = 18.4 cm, d_i = 12.0 cm, h_i = 6.0 cm, M = 0.66
1.5f:
d_0 = 13.8 cm, d_i = 17.2 cm, h_i = 12.0 cm, M = 1.33

At 0.5f (4.6 cm) however, the image appeared too large to see. But when calculated/ predicted, it should have appeared upright.
Since it was not able to be observed directly, the image was only observable if seen through the lens directly. This type of image is virtual. The image, as predicted, was no longer inverted, but upright.

graph of d_i vs d_o

The graph was supposed to be analogous to y=1/x graph, though. 

Q. 8. The y intercept can not be found, because of the hyperbolic nature of the graph. The first graph does not look like to have a y intercept, as (10,27) was the uttermost left point (vertex).

The y value represents the negative inverse of object distance; since it is negative inverse, it represents the virtual distance.

Error and analysis:
The first possible error was during the calculation of the focal point; measuring the triangle's hypotenuse (distance) was done rather loosely; it was done on the hill and not a flat surface. The triangle was no longer perfectly orthogonal anymore.
The second one was observable by looking at the graph; there seemed to be a mistake when recording either the distance or height of the image at 4f. The graph showed a point at y= 36.8 cm (4f), whereas it should have been lower: distance less than the recorded image distance. The magnification then, is somehow off. The graph has two points that were misplaced; it should look like y=k/x, whereas the graph from experiment appeared hyperbolic instead.

1/S + 1/S' = 1/f, whereas S is object distance to lens and S' is the image distance from lens to reflective surface (surface where the reflection is the sharpest).
Solving for f, the following equation was obtained:

A. f = S(S') / (S' + S)
B. M is |S'|/ S, and S' = MS
C. S = S' / M
substituting that into equation A, f = (MS) / (M+1), and substituting C into equation A, f = S' / (M+1)
Focus depends only on either M and S or M' and s.

Conversely, S = S'f / (S' - f); object distance depends on image distance and focus over difference between image distance and focus.
Lastly, using M = |S'| / S, equation M = (S' - f)/ f was obtained. Using S, M= (S-f) / (f * S^2)
Magnification here is dependent on the sum of S and negative f over f times object distance squared. If S - f < 0, meaning object is placed where it is less than focus distance, it will give negative M. The image will be magnified differently. That explains when the object is placed close (S <= 0.5 f) as done in experiment to lens, it would appear upright and magnified and when  it is placed farther than half the focal length, it would appear smaller (0 < M < 1) and inverted.