I'm not very knowledgeable in optics but i'm learning about lens and the thin lens equation:

$\frac1f = \frac{1}{d_o} + \frac{1}{d_i}$

and was wondering about a specific problem that i'm reading.

It's asking me to determine the best focused object distance when a camera is at the same distance of the lens' focal length, 120mm.

$\frac{1}{120} = \frac{1}{d_o} + \frac{1}{120}$

The best focused image object distance would be infinity? What does that actually mean?

Many sites discuss that there are no images formed when the object distance is equal to the focal length, but there arent any I can find that discuss when the image length is equal to the focal lenght

  • $\begingroup$ If the object is placed at the focal plane, then all rays that travel through the lens are parallel. Think about how far the image will form given the rays are parallel. $\endgroup$
    – joseph h
    Commented Feb 7, 2022 at 0:31
  • $\begingroup$ But what about trying to determine the object distance when the image is placed at the focal plane? $\endgroup$
    – dftag
    Commented Feb 7, 2022 at 1:12
  • $\begingroup$ Then the same will be true. You will not be able to define the object distance given that the image forms exactly at the focal plane. If $v=f$ then $\frac1u +\frac1f=\frac1f$ means $\frac1u=0\therefore u\rightarrow \infty$ which is not defined. $\endgroup$
    – joseph h
    Commented Feb 7, 2022 at 2:57
  • $\begingroup$ Since infinity is not very nice to deal with, lets put the image plane almost at the focal plane. say, $d_i =f+\delta$ where $\delta$ is much, much smaller than $f$. Then calculate $d_o$ and see if you can make any sense out of the result. $\endgroup$
    – garyp
    Commented Feb 7, 2022 at 2:59

2 Answers 2


If a camera is placed at the focal point of a lens, the object would have to be placed infinitely far away. This means that rays would have to travel parallel to the optical axis to be in focus at the camera plane.

In practice, it means that if you place the camera arbitrarily close to the focal point, you can focus to objects arbitrarily far away. In the table below I consider a lens with a focal length of 35 mm, which is within the common range of focal lengths for cameras. As you get closer to the focal point, the object distance quickly shoots up. If your camera is not in the image spot provided by the lens equation, your image won't be sharp. But the closer you get, the sharper your image will be. If you set your camera in the focal point of the lens, things that are "very far" away, such as mountains or stars, will be in focus. This is because the error in camera distance quickly becomes neglible, as motivated by the table below.

Note that $0.00001\,\text{mm}=10\,\text{nm}$, which is already smaller than the wavelength of visible light!

$$ \begin{array}{ll} \text{Camera dist.} & \text{Object dist.} \\ \hline 35.1\,\text{mm} & 12.285\,\text{m} \\ 35.05\,\text{mm} & 24.535\,\text{m} \\ 35.02\,\text{mm} & 61.285\,\text{m} \\ 35.01\,\text{mm} & 122.535\,\text{m} \\ 35.001\,\text{mm} & 1.225\,\text{km} \\ 35.0001\,\text{mm} & 12.25\,\text{km} \\ \end{array} $$


You are right in your assumption that the object would need to be at infinity to be at focus at the focal point. Assuming that when they say camera they mean the sensor.

Think starlight which is at infinity but actually it's a 1/d relationship so even for modest distances of say 50 m the image quality won't be much different than for a star.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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