Say we have an object $O$ that is to the left of two lenses. Let's say that the first lens $L_1$ is to the left of $L_2$, the second lens. Assuming we know the type of each lens (concave or convex), the focal length of each lens and the distances between the object and the lenses, we could determine the final image as follows: First, use the lens equation to find the image of the object through $L_1$, and call this image $I_1$. Then you can treat $I_1$ as the object of the second lens $L_2$ by using the distance between $I_1$ and $L_2$. Doing so will produce your final image $I_2$.

What if I decided to go through the same process, but by finding the image that goes through $L_2$ first and then using that image as the object for $L_1$? Does this calculation correspond to a valid image? If not, then why?

Also, I ask the same questions for the case when $L1$ is a lens and $L2$ is a mirror.

I am sorry if this is somewhat basic, but it is bothering me. Thank you for your help in advance!


1 Answer 1


I don't know if it's basic. But you could have done an example to see that they are not the same. IF you wanted to prove that two ways are equivalent, that's hard because you must show that they are the same for all possible cases. However, if you want to check that it is false, you just need one case in which the equality doesn't hold.

So, if you invent any example, you'll easily check that the order does matter.

Now let's see why. The lens equation is

$$ \frac{-1}{a}+\frac{1}{a'}=\frac{1}{f'}$$

with $a=$distance to object; $a'=$ distance to the image; and $f'$ =focal lenght, all of them measured from the origin.

If you look at that equation, you should detect that it is clearly non-linear. It is not a linear combination of things, and it doesn't have linear operators like derivatives. Instead, they are quotients. The inverse frunction $y=1/x$ is clearly non-linear.

As a result, you shouldn't expect any comfortable property, like the sum of solutions being also a solution, nor conmutativity.


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