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Very basic question about optics: suppose I have a thin lens with a given focal length $f$, a screen and an object at distance $a$ from the screen. What is the mathematical relation that must be fulfilled so that the image of the object is sharp on the screen?

Thank you very much in advance for your answers.

Julien.

Pxx
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1 Answers1

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Use $${1\over x}+{1\over a-x}={1\over f}$$

Qwerty
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  • When both distances are equal, then each distance = $$ 2* f ; ,1/ (\alpha) + 1/(\alpha) = 1/f ;, 1/ (2f) + 1/(2 f) = 1/f $$ – Narasimham Nov 25 '16 at 18:03
  • Why is this case special?Has the OP asked for it? – Qwerty Nov 25 '16 at 18:06
  • Special case because $u=v=a $ as stated by OP.Normally $u \ne v$. Yes OP asked for it. – Narasimham Nov 25 '16 at 18:52
  • @Narasimham Either I or you are misunderstanding. I think the object is at a distance $a$ from the screen i.e. $u+v=a$ and not $u=v=a$ – Qwerty Nov 25 '16 at 19:14
  • I understood his question as screen (image) distance $u = a $ = object distance $=v =a$. – Narasimham Nov 25 '16 at 19:23
  • @Narasimham But object distance is the distance of the object from the lens and not from the screen. Similarly, image distance is the distance of the screen from the lens, not the object – Qwerty Nov 25 '16 at 19:29
  • Right, I am talking about answer, not question. OP seems to confuse the two. Bye.. – Narasimham Nov 25 '16 at 19:36
  • Thank you @Qwerty that's exactly what I needed. Narasimham thank you for your answer too but you misunderstood my question, and I don't think I'm being confused between distance lens-object and screen-object. That $a$ is the distance object-screen was clearly stated in my question. – Pxx Nov 25 '16 at 21:02
  • I have an additional question, if I may @Qwerty. If I were to place a second identical lens (so same focal length $f$) at a distance $d_a$ from the 1st lens in direction of the screen, could I say that the image is sharp if the following relation holds: $$\frac{1}{f} = \frac{1}{x} + \frac{1}{d_a} = \frac{1}{d_a} + \frac{1}{a-x-d_a}$$ The idea is that the image must reach the 2nd lens sharply (1st equality) first, but I might be missing a factor $\frac{1}{2}$ somewhere (because of the location of the focal point between the lenses?). – Pxx Nov 25 '16 at 21:08
  • @Julien As far as I remember, there is a formula(I don't remember it now) that converts $2$ lenses into one lens , with appropriate positioning. – Qwerty Nov 26 '16 at 04:46