Answer
$19.01\;\rm cm$
Work Step by Step
We know that the image from the first lens (the diverging one) is the object of the second lens (the converging one).
So, we need to find the object length of the second lens $d_{o2}$ which is the image of the first lens.
$$\dfrac{1}{f_2}=\dfrac{1}{d_{o2}}+\dfrac{1}{d_{i2}}$$
Solving for $d_{o2}$;
$$\dfrac{1}{f_2}-\dfrac{1}{d_{i2}}=\dfrac{1}{d_{o2}} $$
$$d_{o2}=\left[\dfrac{1}{f_2}-\dfrac{1}{d_{i2}} \right]^{-1}$$
Plugging the given;
$$d_{o2}=\left[\dfrac{1}{12}-\dfrac{1}{17} \right]^{-1}=\bf 40.8\;\rm cm$$
Now we can find the image distance of the first lens
$$d_{i1}=40.8-30=10.8\;\rm cm$$
and since the distance between the two lenses is 30 cm, so the image is on the same side as the object which means it is a virtual image.
Thus, $d_{i1}=\bf -10.8\;\rm cm$
Now we can find the focal length of the first lens.
$$f_1=\left[\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}}\right]^{-1}$$
Plugging the known;
$$f_1=\left[\dfrac{1}{ 25}+\dfrac{1}{-10.8}\right]^{-1}=\bf -19.01\;\rm cm$$
Therefore, the focal length of the diverging lens is 19 cm.
