## Day 2.26 Lens Lab Data – Call for Help

Thurs. Mar 6, 2014

Physics 2 students were given the freedom to investigate anything they felt compelled to using a light source, converging lens, and viewing screen.  Some groups choose to measure the distance from the light to lens and compare to the distance from the lens to image.  This was good, and the data modeled the Lens Equation  (1/do + 1/di = 1/f ) perfectly!

The above group (and one other one) choose to measure the distance from light to lens vs. the size (height) of the image produced.  The board above shows their data.  The raw data shows an inverse relationship.  So we decided to graph y vs. 1/x, this straightened the data out, so good, right?  Now comes the tough part.  What is the data modeling for us?  Here is where we are stuck.  One thing to keep in mind the focal length of the lens used is 15.0 cm.  (BTW, the other group used the same focal length lens and ended up with similar data.)

do = (14.67cm^2)(hi) + 14.91 cm

From what I can figure the y-intercept is the focal length of the lens, because you would get a image of size 0cm if the light was at the focal length of the lens.  BUT, what is the slope indicating?  What is with the units of cm^2?  Is it a coincidence that this number is also close to the focal length, or is it something significant?  What are we missing here.  Any and all help will be appreciated.  Please comment below or tweet me @MrBWysocki.

Thanks for the help!

I teach high school Chemistry and Physics using Arizona State University's Modeling Method. The school I teach in is Bloomer HS in Northwest Wisconsin. I've been using the Modeling Method for 7 years now and have just recently begun using Standards Based Grading as a way to assess my students learning.
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### 3 Responses to Day 2.26 Lens Lab Data – Call for Help

1. Scott Hertting says:

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.

2. Pippa says:

Hi Mr Wysocki,

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!