A concave lens is in contact with convex lens. The ratio of the magnitude of their power is $2/3$ . Their equivalent focal length is$30cm.$  What are their individual focus lengths in $cm$ ?

Let $f_1$ be the focal length of the concave lens and $f_2$ be the focal length of convex lens. The power of lens is negative for concave lens and positive for convex lens. i.e. $P_1$ is negative and $P_2$ is positive. We know that power$\left(P\right)=1/f$ .

So, as per question, $\frac{P_1}{P_2}=\frac{f_2}{f_1}=-\frac{2}{3}$ or$f_2=-\frac{2}{3}{f}_1$

And $\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{30}$ or$-\frac{3}{2f_1}+\frac{1}{f_1}=\frac{1}{30}$

On solving, $f_1=-15cm$ and$f_2=10cm.$