# The image of a wall clock is to be obtained on the opposite wall 2m away by the means of a convex lens. What is the minimum focal length required? [closed]

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?

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

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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 $d/4=0.5\mathrm{m}$.