Let the focal length of the convex lens be \(f_{1}\), and that of the concave lens be \(f_{2}\).
According to the question, we have the focal length of the convex lens equal to \(10 {~cm}\), and that of the concave lens equal to \(20 {~cm}\).
We know from the Cartesian sign convention that the focal length of a convex lens is positive while that of a concave lens is negative.
So the values of the focal length with proper sign convention are given as:
\( f_{1}=+10 {~cm}\quad\ldots(1)\)
\(f_{2}=-20 {~cm}\quad\ldots(2)\)
Now, we know that the equivalent focal length of the combination of thin lenses in contact is given by the formula,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}+\ldots \ldots+\frac{1}{f_{n}}\)
As there are two lenses of focal lengths \(f_{1}\) and \(f_{2}\), so the equivalent focal length of the given combination is given by,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
Substituting equation (1) and (2) we have:
\(\frac{1}{f}=\frac{1}{+10}+\frac{1}{-20}\)
\(\Rightarrow \frac{1}{f}=\frac{2-1}{20}\)
\(\Rightarrow \frac{1}{f}=\frac{1}{20}\)
Taking the reciprocal we get,
\(f=+20 {~cm}\)
Thus, the focal length of the given combination of the lenses is equal to \(+20 {~cm}\).
Hence, the correct option is (B).