What is the actual focal length?

[argarg03]

I understand completely how to go about doing this problem... I know that you would have to plot the points and draw a line of best fit and that the x-intercepts and y-intercepts would equal 1/f... but what would the actual focal length be? and how do you calculate the gradient? This is actually a review sheet for my test and the teacher didn't give us the answer to this last one because it was the last thing we covered so he didn't get the chance to. I would really like to know the exact answer so I can review it multiple times as the test is very similar to these problems.

It says: A converging lens is set up on an optical bench and the distances to the object O and image I are measured. Use a graph to determine the focal length of the lens.

O,cm I,cm
20 80
25 44.1
46 24.5
58 22.1
70 20.7

Thanks so much in advance!

 

[zachzach]

I understand completely how to go about doing this problem... I know that you would have to plot the points and draw a line of best fit and that the x-intercepts and y-intercepts would equal 1/f... but what would the actual focal length be? and how do you calculate the gradient? This is actually a review sheet for my test and the teacher didn't give us the answer to this last one because it was the last thing we covered so he didn't get the chance to. I would really like to know the exact answer so I can review it multiple times as the test is very similar to these problems.

It says: A converging lens is set up on an optical bench and the distances to the object O and image I are measured. Use a graph to determine the focal length of the lens.

O,cm I,cm
20 80
25 44.1
46 24.5
58 22.1
70 20.7

Thanks so much in advance!

Use the lens equation for each of the data points:

[tex] \frac{1}{f} = \frac{1}{O} + \frac{1}{I} [/tex]

 

[zachzach]

Ohh it says use a graph! DOH!

 

[zachzach]

So graph O on the x-axis and I on the y-axis and solve the lens equation for I.

 

[argarg03]

but I'm solving for focal length... for another example the number on the intercepts was 0.14

would that be the focal length or would I actually have to solve 1/.14 which comes out to like 7.0 something... or would the focal length be just 0.14

Is this confusing? I think I'm confusing myself

 

[zachzach]

but I'm solving for focal length... for another example the number on the intercepts was 0.14

would that be the focal length or would I actually have to solve 1/.14 which comes out to like 7.0 something... or would the focal length be just 0.14

Is this confusing? I think I'm confusing myself

I just graphed it like I said and there are no intercepts. The focal length is neither of those though. it is semi-confusing...just need to think about it. I know you are solving for f, but your given data points of O and I, so just make x = O and y = I and then somehow you will be able to solve for f from the graph.

 

[zachzach]

But if you have an example then i must be graphing it wrong.

 

[zachzach]

O..ok just graph (1/O) on the x-axis and (1/I) on the y-axis so then your equation is in the form y = mx+ b.

 

[zachzach]

So then the slope would be -1 and b = (1/f).

 

[zachzach]

[tex]

\frac{1}{I} = -1\frac{1}{O} + \frac {1}{f} [/tex]
[tex]

y \,= \;\; mx \;+ \:b \;\;\;\rightarrow \;\;b = \frac{1}{f}[/tex]