So yesterday, we did a practical in Physics, where we found the focal length of a convex lens, through an experiment where we had 2 needle-like objects and a convex lens placed between them, and we had to adjust the lens in a certain manner. Since the image of the object (say A) would be inverted, it's needle would appear to point downward in the lens and we had to adjust lens such that the pointed end of the image that we see of object A in the lens must co-incide with object B's pointed end.

But I never understood why we are doing that? I know that we require 2 variables in determining the focal length, one is (v) the image distance from the lens and the other is (u) the object distance from the lens.

I have a lot of confusions about this experiment so I'll start with the most basic one.

  1. Why are we able to see the real image in the lens of a convex lens? Why does it appear to be inside the convex lens, I understand why we are able to see a virtual image formed by a concave lens as being formed inside the lens, as when we trace the light rays back, we appear to get an image behind the lens.

But since the image that is formed by a convex lens is real and intersects at a point in front of the lens, why is it that I am able to see the inverted image of it being formed inside the lens? Shouldn't I need a screen to obtain the image, and if the retina of my eyes is acting as a screen here then why do I see it being formed inside the lens?

2)So another mystery that seems to me is why do we have another needle (Object B) and we have to place it in a certain manner (as described above) and that distance that we measure of Object B when it is placed in that certain manner is the distance of the image that is formed by the convex lens of Object A? I am just not able to grasp how that would be.

I hope someone can understand my query and resolve it.


1 Answer 1


In the method that you're using you need to make two measurements because you are using the relation $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ so you need to measure the object and image distances, $d_o$ and $d_i$. Another common technique for measuring the focal length is to find the image distance for an object that is distant enough that the object distance can be considered infinite. In that case $f=d_i$.

To answer why you can see a real image it helps to remember that images look like objects because light comes from an images in the same way as it comes from an object. The real image is formed in front of the lens, that is to say on the same side of the lens as you and the opposite side compared to object A. Rays of light from the object pass through the lens, converge at the point of the image and then diverge again. You speak of seeing the image "in the lens" and it's true the image has to be lined between your eye and the lens but if you look carefully and move you head slightly while looking at the image you will find that it behaves as if it were in front of the lens, rather behind it as in a virtual image.

You could hold a screen (piece of paper) at the point of where the image is formed and you might be able to see the real image projected on it though probably not. This would work more reliably if you were using a candle or some other luminous object. With respect to the retina of you eye, what is happening there is that your eye's lens (with the cornea) is forming a real image on your retina. This process is the same regardless of whether you are looking at a real image, or a virtual image or indeed the original object itself.

In the lab procedure you are trying to place object B so that it coincides with the image of object A. The idea is if get it in the right spot you can move your head around a bit and the object A's image and object B should appear to remain together. If they seem to move relative to each other as you change your perspective then you know that they are not yet in the same place.

This may seem like a complicated way to measure the focal length but it does give you some experience of working with real images.


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.