Answer
-26.7 cm.
Work Step by Step
The image formed by the converging lens is 30.0 cm from the converging lens. This becomes a virtual object for the diverging lens, which is placed 15.0 cm to the right of the diverging lens. Therefore, for the diverging lens, s = -15.0 cm.
The final image is a distance of 15.0cm+19.2cm=34.2cm away, so s’= 34.2 cm.
Use the lens equation.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
$$\frac{1}{-15.0cm}+\frac{1}{34.2cm}=\frac{1}{f}$$
$$f=-26.7cm$$
