In the following derivation of combination of thin lenses, why no sign convention is applied? For the first step, image distance, $u=+v_e$, but shouldn't it be $-v_e$ as the object is to the left of the lens? Also, in the second step, since the virtual object is to the right of the lens, shouldn't both the image distance, $v$ and object distance, $v1$ be $-v_e$?

Combination of Thin Lenses in contact

Let us consider two lenses $A$ and $B$ of focal length $f_1$ and $f_2$ placed in contact with each other. An object is placed at $O$ beyond the focus of the first lens $A$ on the common principal axis. The lens $A$ produces an image at $I_1$. This image $I_1$ acts as the object for the second lens $B$. The final image is produced at $I$ as shown in figure. Since the lenses are thin, a common optical centre $P$ is chosen.

Let $PO = u$, object distance for the first lens ($A$), $PI = v$, final image distance and $PI_1 = v_1$, image distance for the first lens ($A$) and also object distance for second lens ($B$).

Combination of Thin LensesFor the image $I_1$ produced by the first lens $A$,

$$\frac{1}{v_1} – \frac{1}{u} = \frac{1}{f_1} \tag{1}$$

For the final image $I$, produced by the second lens $B$,

$$\frac{1}{v} – \frac{1}{v_1} = \frac{1}{f_2} \tag{2}$$

Adding equations (1) and (2),

$$\frac{1}{v} – \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \tag{3}$$

If the combination is replaced by a single lens of focal length $F$ such that it forms the image of $O$ at the same position $I$, then

$$\frac{1}{v} – \frac{1}{u} = \frac{1}{F} \tag{4}$$

From equations (3) and (4),

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \tag{5}$$

This F is the focal length of the equivalent lens for the combination. enter image description here

  • $\begingroup$ It's just conventions. Different authors have different preferences. Whichever the convention used, the result will be the same if it's applied correctly. I like the one where object distances to the left and image distances to the right are positive, but up is always positive and down is always negative. $\endgroup$ Commented Jan 25, 2020 at 7:52
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    $\begingroup$ @relatively_random But then the result wouldn't be the same if we apply the sign convention as I have suggested in the question. $\endgroup$ Commented Jan 26, 2020 at 6:29

1 Answer 1


The author did use a convention, probably just not the one you're used to. From the form of the well-known thin lens equation used ($\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$), we can conclude that the author considers both object and image distances positive towards the right and negative towards the left. It's a simple convention that many people like.

With the derivation you posted, it's easy to forget that $v_1$ (the image of the first lens) and $u_2$ (the object of the second lens) are not the same. In this case, they both have the same value because the lenses are in contact (zero distance between them) and because both objects and images are positive in the same direction with the convention the author used. So for the second lens, we can use $u_2 = v_1$ and $v_2 = v$ to turn $\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$ into $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$.

Personally, I prefer the convention where object distances are positive towards the left and image distances positive towards the right. To my engineering brain it just seems more natural, and the thin lens equation is then nicely positive all over: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. I admit it does complicate things sometimes, but at least it's more difficult to just blindly plug the same numbers somewhere they don't logically belong.

With the convention I prefer, $u_2 = -v_1$ (because the images and objects are positive in opposite directions) and $v_2 = v$, so the equation for the second lens $\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f_2}$ becomes $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$. Here's the derivation repeated with this convention:

$$\frac{1}{v_1}+\frac{1}{u}=\frac{1}{f_1}$$ $$\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$$

Adding them together: $$\frac{1}{u}+\frac{1}{v}=\frac{1}{f_1} + \frac{1}{f_2}$$

And substituting this full compound lens equation: $$\frac{1}{u}+\frac{1}{v}=\frac{1}{F}$$

Gets us the same result: $$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$$


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