Step 1: Identify focal length of diverging lens.
For a diverging lens:
\[
f_1=-20\,\text{cm}
\]
Incident rays are parallel, so the diverging lens forms a virtual image at its focus.
\[
v_1=-20\,\text{cm}
\]
Step 2: Locate this image with respect to second lens.
The converging lens is placed \(10\) cm to the right of the diverging lens.
The image formed by the first lens is \(20\) cm to the left of the diverging lens.
So distance of this image from second lens:
\[
20+10=30\,\text{cm}
\]
Thus for the second lens:
\[
u=-30\,\text{cm}
\]
Step 3: Write focal length of converging lens.
For converging lens:
\[
f_2=+15\,\text{cm}
\]
Step 4: Apply lens formula.
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
\[
\frac{1}{15}=\frac{1}{v}-\frac{1}{-30}
\]
\[
\frac{1}{15}=\frac{1}{v}+\frac{1}{30}
\]
Step 5: Solve for \(v\).
\[
\frac{1}{v}=\frac{1}{15}-\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{2-1}{30}
\]
\[
\frac{1}{v}=\frac{1}{30}
\]
\[
v=30\,\text{cm}
\]
Step 6: Interpret the sign of image distance.
Since \(v\) is positive, the final image is formed to the right of the converging lens.
A positive image distance for the second lens indicates a real image.
Step 7: Final conclusion.
\[
\boxed{\text{Real image at }30\,\text{cm to the right of the converging lens}}
\]