Q.A convex lens of focal length $+15\ \text{cm}$ is placed in contact with a concave lens of focal length $-25\ \text{cm}$. Find the power and focal length of the combination, and state whether the combined system converges or diverges a beam of parallel light.

$P_1=100/15\approx6.667\ \text{D}$ (convex, converging, positive) and $P_2=-100/25=-4\ \text{D}$ (concave, diverging, negative). Adding, $P=P_1+P_2=6.667-4=2.667\ \text{D}$. The equivalent focal length is $F=100/P=100/2.667\approx37.5\ \text{cm}$. Because the resultant power is positive (the stronger convex lens's converging effect outweighs the weaker concave lens's diverging effect), the combination as a whole still behaves as a converging system, though a much weaker one than the convex lens alone ($37.5\ \text{cm}$ versus $15\ \text{cm}$).

Convert each focal length to power (with the correct sign for convex/concave), add the powers, then invert to get the equivalent focal length.

• Forgetting the concave lens's power must be entered as negative before adding.
• Concluding the system diverges just because one of the two lenses is diverging — the SIGN of the sum, not of either lens alone, decides.