1
$\begingroup$

I'm in 10th grade and this question came in my physics test. Nobody was able to answer this question correctly except my physics teacher who says that the answer is 2m. My answer is that there should be no limit on how small the focal length needs to be in this case.

For example, if a convex lens of focal length 2cm is used to form an image on the opposite wall, the wall clock that is 2m away from the lens can be treated to be at infinity with respect to the lens and if the focus of the lens is kept at the opposite wall, an image of the object at infinity should form on the wall. My argument is that no matter how small the focal length becomes, as long as it's above zero, an image should form on the opposite wall. Please try to solve the problem and post the explanation.


edit:

I asked my teacher and he told me that while solving the equation through the lens formula, he had taken the object position to be infinite and the image distance at 2 meter.

If there were a minimum limit on how small the focal length could be, what would it be?

$\endgroup$

closed as off-topic by Kyle Kanos, John Duffield, JamalS, user36790, ACuriousMind Feb 14 '16 at 18:49

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" – Kyle Kanos, John Duffield, JamalS, Community, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

We have that $1/f=1/b+1/g$, where $f$ is the focal length and $b$ and $g$ are the distance of object and screen to the lens, respectively. Thus, $$ \frac{1}{f}=\frac{1}{b}+\frac{1}{g}=\frac{b+g}{bg}\ . $$ Using that $b+g=d=2\mathrm{m}$, it follows that $f=bg/d=b(d-b)/d$. It is easy to see that $b(d-b)$ takes values between $0$ and $d^2/4$. The minimal value of $f$ is thus zero, and the maximum $d/4=0.5\mathrm{m}$.

$\endgroup$
1
$\begingroup$

If you have a convex lens of focal length f the minimum distance which is possible between object and real image is 4f. So it would appear that any convex lens with a focal length of less than or equal to 50 cm will do. It also shows that the answer is not 2 metres.

$\endgroup$
  • $\begingroup$ Doesn't this mean that 50 cm is the maximum possible focal length and not the minimum focal length required to form a real image? $\endgroup$ – Aniansh Feb 13 '16 at 18:02
  • $\begingroup$ Yes. For a focal length of 50 cm the object and image distances would be both 1 metre. $\endgroup$ – Farcher Feb 13 '16 at 18:26
  • $\begingroup$ @Aniansh Have you edited your original question because I do not think that what is written at present was the question that you originally asked? In the original question the implication was that the clock was 2 metres from the wall. $\endgroup$ – Farcher Feb 14 '16 at 12:43
  • $\begingroup$ the question is the same and your answer is also good but what i want to know is wether there is a minimum limit of the focal length required or not. I want to know how small can the focal length be in this case. $\endgroup$ – Aniansh Feb 14 '16 at 13:10

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