A convex lens and a concave lens are placed in contact. The ratio of the magnitude of the power of the convex lens to that of the concave lens is $3 : 2$. If the focal length of the concave lens is $10 cm$, the focal length of the combination is

Let $f_1$ be the focal length of the convex lens and $f_2$ that of the concave lens. Given $f_2=-10 cm$ and

$\left|\frac{P_1}{P_2}\right|=\frac{3}{2} or \frac{f_2}{f_1}=\frac{3}{2}$

Since $f_2$ is negative,  $$\begin{gathered} \frac{f_2}{f_1}=-\frac{3}{2} \\ {f}_1=-\frac{2}{3}{f}_2 \end{gathered}$$

$$\begin{gathered} f_1=-\frac{2}{3}\times -10 \\ =\frac{20}{3} cm \end{gathered}$$

 The focal length of the combination is given by

$$\begin{gathered} \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \\ =\frac{3}{20}-\frac{1}{10} \\ =\frac{1}{20} \end{gathered}$$

which gives $F = 20 cm$.