Two thin convex lenses have focal lengths of 24 cm and 16 cm…

Step 1. The equivalent focal length. $\dfrac{1}{F} = \dfrac{1}{24\ \text{cm}} + \dfrac{1}{16\ \text{cm}} - \dfrac{8.0\ \text{cm}}{(24\ \text{cm})(16\ \text{cm})} = 0.04167 + 0.06250 - 0.02083 = 0.08333\ \text{cm}^{-1}$. So $F = 12.0\ \text{cm}$.

Step 2. The power. $P = \dfrac{1}{0.120\ \text{m}} = 8.33\ \text{D}$.

Step 3. The focus. $F\left(1 - \dfrac{d}{f_1}\right) = 12.0\ \text{cm} \times \left(1 - \dfrac{8.0}{24}\right) = 8.0\ \text{cm}$ behind the second lens.