We use the lens equation
$\begin{align*}\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\end{align*}$
where $d_o$ is the distance from the object to the lens, $d_i$ is the distance from the image to the lens, and $f$ is focal length. This equation can be used with two lenses by first applying it to find the first lens, ignoring the second lens, and then applying it to the second lens, ignoring the first lens, and taking the image of the first lens as the object of the second lens.

The conventions for the lens equation
tell us that the focal length is positive for a converging lens and negative for a diverging lens. Since both of the lenses in this problem are converging, the focal lengths are $f$ are $20 \mbox{ cm}$ and $10 \mbox{ cm}$ , respectively. The object distance for the first lens is positive, because the object is on the same side of the lens from which the light is coming. Thus, the lens equation for the first lens is:

$\begin{align*}\frac{1}{40 \mbox{ cm}} + \frac{1}{d_i} = \frac{1}{20 \mbox{ cm}} \Rightarrow d_i = 40 \mbox{ cm}\end{align*}$
This is $10 \mbox{ cm}$ behind the second lens. Therefore, $d_o$ in the lens equation for the second lens is $-10 \mbox{ cm}$ . Thus, the lens equation for the second lens is
$\begin{align*}\frac{1}{-10 \mbox{ cm}} + \frac{1}{d_i} = \frac{1}{10 \mbox{ cm}} \Rightarrow d_i = 5 \mbox{ cm}\end{align*}$
The image distance is positive when the image is on the opposite side of the lens from which light is coming. Therefore, the image is $5 \mbox{ cm}$ to the right of the second lens. Therefore, answer (A) is correct.

Solution to 2001 Problem 11