I think I might have an answer for you.

Let’s say (I’m assuming this is the terminology you’re using):

do = distance from lens to light

di = distance from light to image

ho = height (size) of object

hi = height (size)

Using the magnification equation (see, e.g. http://en.wikipedia.org/wiki/Lens_(optics)#Imaging_properties), which states that the magnification factor (amount the image is bigger/smaller than the object) M is:

M = – di / do = f / (f – do)

M must also equal hi / ho. (From logic: the ratio of one height to the other must be the amount that the image is bigger than the object).

Therefore, hi / ho = f / (f – do). Rearranging (multiplying up both fractions to the opposite sides):

hi (f – do) = ho x f

hi x do = hi x f – ho x f

do = f – (f x ho) x 1 / hi (your x-axis term is 1 / hi)

Bearing in mind that, presumably, one of your image heights is negative (i.e the object’s height is measured below the central focal line and the image’s height above, or vice versa), the signs here will cancel out, so you’ll get:

do = (f x ho) x 1 / hi + f

This gives you your intercept term of f (14.91 is roughly equal to 15), and suggests that the gradient you’re seeing is the focal length, 15cm, multiplied by the height of the object (a quick reverse calculation suggests the height of the object was 14.67/15 = 0.98 cm). The cm^2 therefore comes from the multiplication of the two.

There isn’t a coincidence as such: it’s just that the object height was close to 1, therefore the focal length dominated the gradient term.

Definitely worth checking my maths and seeing if the calculation for the object height was right, though!

]]>The un-linearized data seems to make sense from a conceptual standpoint. I’m going to guess we are just focusing (no pun intended) on real images. As the object gets farther from the lens (so do increases) AND we are way outside 2f, the image will be smaller and keep getting smaller. But I think the size of the image would have to approach the object size?Now I’m not totally sure about this. Here is my thinking… As do increases di decreases and approaches the focal length(think about the usual di vs. do lab). So di ~ f. Now go to the ‘similar triangle’s equation’ and solve for do: do= (ho/hi)*di. Rearrange to do= (ho*di)*1/hi, but di~ f so do= (ho*f) *1/hi, now it is in the form of your graph (y=mx+b) with do on the y-axis and 1/hi on the x-axis. In my mind the slope of the graph always relates to something help constant in the experiment, in this case ho and f. But I’m not sure that fits unless your object was about 1.0cm tall with your 15cm focal length.

As for the intercept? No idea at this point. I’m probably way off in above anyway. Really interesting question/results. ]]>