Answer
$\frac{f_1f_2}{f_1+f_2}$
Work Step by Step
We know that for the image formed by the lens A, we have
$\frac{1}{d_i}-\frac{1}{d_{\circ}}=\frac{1}{f_1}$......eq(1)
and for the image formed by the second lens B, we have
$\frac{1}{d_i^{\prime}}-\frac{1}{d_i}=\frac{1}{f_2}$......eq(2)
Adding eq(1) and eq(2), we obtain:
$\frac{1}{d_i^{\prime}}-\frac{1}{d_{\circ}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\implies \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
$\implies \frac{1}{f}=\frac{f_1+f_2}{f_1f_2}$
$\implies f=\frac{f_1f_2}{f_1+f_2}$
This is the required effective focal length.
