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I am trying to understand how thin lens camera work. From what I understand of thins lens model, the image is formed on image plane that obey the equation

$$\frac1S_1 + \frac1S_2 = \frac1f$$

enter image description here

So how about the case when there are 2 Object with different sizes and the distances of two object to the lens are difference while the image plane of the camera is fixed. Also, the two objects and the lens don't form a straight line so there is no way that one object hide the other one. Let say object A and B has distance $S_A$ and $S_B$. Suppose that

$$\frac1S_A + \frac1S_2 = \frac1f$$

So in the case of $S_B$

$$\frac1S_B + \frac1S_2 \neq \frac1f$$

Then how the image of object B is form on image plane of camera

Assume that both object have their distance from the lens greater than focal length.

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    $\begingroup$ You just found out that a camera lens focused at one object will create blurry images of objects at other distances. $\endgroup$ – Han-Kwang Nienhuys May 25 '16 at 22:12
  • $\begingroup$ How does the image of object B form on the image plane? It doesn't. If you perceive that it does, that's either because you have a large depth of focus, or you are tolerant of the blurriness that results. $\endgroup$ – garyp May 25 '16 at 22:34
  • $\begingroup$ Perhaps the Wikipedia article "Depth of field" will help? en.m.wikipedia.org/wiki/Depth_of_field $\endgroup$ – Farcher May 26 '16 at 4:47
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The equation holds true to all object, no matter their distance from the lens, HOWEVER when f is fixed, and the object are at different distances from the lens, then it follows that if one of the object is forming an image, then the other must not!

If you had a lens with fixed focal length, and two object at different distances from it, but now you have a screen you can move closer and further from the lens (instead of a fixed sensor), you would find out that both objects can be imaged by the lens - just at different distances from the lens, according to the equation.

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