1
$\begingroup$

An older camera has a lens with a focal length of 50mm and uses 34-mm-wide film to record its images. Using this camera, a photographer takes a picture of the Golden Gate Bridge that completely spans the width of the film. Now he wants to take a picture of the bridge using his digital camera with its 12-mm-wide CCD detector. What focal length should this camera's lens have for the image of the bridge to cover the entire detector?

I figured out that I can construct two equations using the lens equation. Each equation corresponds to each case (1) the film and (2) CCD.

Moreover, we need to use the length-distance relationship in order to express the distance term in the lens equation in terms of the length of the image.

$$ m = \frac{y_o}{y_i}=\frac{d_o}{d_i} $$

where $y_o$ and $y_i$ are the lengths of the object and the image, and $d_o$ and $d_i$ the distances of the object and the image from the lens.

For the film, we have $$ \frac{y_o}{34}=\frac{d_o}{d_i} \Rightarrow d_i = \frac{34d_0}{y_0} $$ and for the CCD $$ \frac{y_o}{12}=\frac{d_o}{d_i} \Rightarrow d_i = \frac{12d_0}{y_0} $$

Hence we have the equations $$ \frac{1}{d_o} + \frac{y_0}{34d_o} = \frac{1}{50} $$ $$ \frac{1}{d_o} + \frac{y_0}{12d_o} = \frac{1}{f} $$

I think we can only set up these two equations as far as the problem is concerned. The difficult part is dealing with three variables in just two equations. I wonder if I have set up the lens equation properly or if the equations can be simplified further.

$\endgroup$

closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind, Jim, BMS Feb 25 '15 at 20:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, ACuriousMind, Jim, BMS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ I believe I did my work on this problem. Do you have any suggestions on improving this question? This example homework question physics.stackexchange.com/questions/16182/… is considered a good one and I don't think I did less than the person who posted this question. $\endgroup$ – akino Feb 26 '15 at 1:06
  • $\begingroup$ First, pulling a question from 4 years ago isn't a good comparison because site policies change over the years. Second, your question is asking us to check your work for you, something that is considered off topic. Third, removing the HW question from the linked Q leaves a decent question behind; removing the HW question here leaves nothing behind (in my opinion) which makes it a bad question. $\endgroup$ – Kyle Kanos Feb 26 '15 at 3:17
2
$\begingroup$

I think the key point that you are missing is that the object is very far away compared to all of the other distances. Combine your final two equations to eliminate $y_0$ and then let $d_0\rightarrow\infty$, i.e. $\tfrac{1}{d_0}\rightarrow0$.

$\endgroup$
  • $\begingroup$ Thanks for pointing this out. My equations are valid after all. Is this type of question not allowed on here? I believe I did my work. It's sad that 5 people tagged this as off-topic and I'm in danger of being blocked of asking questions in the future. $\endgroup$ – akino Feb 26 '15 at 2:16
  • $\begingroup$ @akino I personally think your question is fine. As you've pointed out, you didn't post a mindless 'do my work for me' question which is what is truly not allowed here. Those users probably voted to close the question because it is about a specific problem instead of a 'concept'. Physics.SE is a big community, and it only takes 5 people to close a question. Don't be discouraged by it. :) $\endgroup$ – Chris Mueller Feb 26 '15 at 3:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.