Step 1: Argue using power instead of focal length.
The power of a lens is \(P=\dfrac{1}{f}\) (with \(f\) in metres), and for lenses in contact the powers simply add: \(P=P_{1}+P_{2}\). This is often the quickest way to combine lenses.
Step 2: Write each power with its sign.
The convex lens has power \(P_{1}=+\dfrac{1}{f}\) and the concave lens, of equal focal-length magnitude, has power \(P_{2}=-\dfrac{1}{f}\).
Step 3: Add the powers.
\[ P=P_{1}+P_{2}=+\frac{1}{f}-\frac{1}{f}=0 \]
Step 4: Convert back to focal length.
Since \(P=\dfrac{1}{F}=0\), the equivalent focal length is
\[ F=\frac{1}{P}=\frac{1}{0}=\infty \]
A zero-power, infinite-focal-length system neither converges nor diverges the beam, so it acts just like a flat glass slab.
Result:
\[\boxed{P=0,\ F=\infty}\]