1
$\begingroup$

enter image description here

the focal length is the distance between the center of the convex lens and the focal point.

in the image illustrated above, we obtain focused image at the (position of image, inverted yellow arrow) not at the focal point.

to determine the focal length of the lens practically, we change the position of the image until we obtain a clear image which will be at the yellow arrow (not at focal point).

so my question is how this method is true since the image will not be formed at the focal point, instead it will be beyond the focal point,therefore the result of focal length obtained will be false, can anyone help me please?

$\endgroup$
  • $\begingroup$ Nice picture! Did you draw it yourself? $\endgroup$ – Floris Mar 6 at 20:30
  • $\begingroup$ Use ceiling light (basically an object at infinity) and then try to form an image of the ceiling. This is how I find focal length of old, unlabelled thin lenses in the lab. $\endgroup$ – wcc Mar 6 at 21:36
1
$\begingroup$

One can also use Bessel's method, moving the lens between the two positions where there is a focused image (enlarged or smaller): $$f = \frac{D^2 -d^2}{4D},$$ where $D$ is the distance between object and image and $d$ the distance between the two ens positions.

This also works for thick lenses.

$\endgroup$
0
$\begingroup$

The lensmaker's equation, $\frac {1}{p} + \frac {1}{q} = \frac {1}{f}$ is all you need if you have p and q. Just solve it for $f$, which is the focal length (the distance from the lens to the focal point).

$\endgroup$
  • $\begingroup$ I know this equation, but how they apply it since the image will not be formed at the focal point!! $\endgroup$ – Rami ki Mar 6 at 18:25
  • $\begingroup$ @Ramiki The equation is a good approximation for the physical process which occurs. "Focal Point" is simply a name for a location. You measure $p$ and $q$ and calculate $f$. $\endgroup$ – Bill N Mar 6 at 18:40
0
$\begingroup$

Place the object very far away from the lens, so that the object distance is as great as possible. The larger the object distance, the more insignificant is the difference between focal length and image distance.

Better is to use a source of light that resembles a point source, such as the tip of an optical fiber, and place this at the focal length of a second lens so that a collimated beam is produced. Use the lens you are testing to focus the collimated beam to a point. The distance from the lens to the focus is the focal distance.

$\endgroup$
0
$\begingroup$

The focal point is a property of the lens. It is where parallel light would converge (in practice, that is light that "appears to come from infinitely far away").

As others pointed out, the lensmaker's formula describes how the object $p$ and image $q$ distance are related to this property of the lens. So you just measure $p$ and $q$, and find $f$ from

$$f = \frac{p\cdot q}{p+q}$$

(I took the liberty of rearranging the usual formulation $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$ so you obtain $f$ directly from your measured values).

We call it the focal point - but most of the time it's not where the image is located. Once you understand you are dealing with a property of the lens, not of the imaging scenario, I think this will be clear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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