Explanation:
The powers add algebraically: P = 1/0.20 − 1/0.10 = 5 − 10 = −5 D. Hence f_eq = 1/P = −0.20 m = −20 cm, so the combination is diverging.

Explanation:
The required equivalent power is −4 D, while the converging lens has power +10 D. Therefore, the diverging lens must have power −14 D, giving f = 1/(−14) m = −0.0714 m = −7.14 cm.

Explanation:
Their powers are −5 D and −3.33 D. Thus, P_eq = −8.33 D and f_eq = 1/(−8.33) m ≈ −0.12 m = −12 cm.

Explanation:
The convex lens has power +4 D. For the resultant power to be negative, the concave lens must have power less than −4 D in algebraic value, meaning its magnitude must exceed 4 D. Therefore, its focal-length magnitude must be less than 25 cm.

Explanation:
The equivalent power is −2 D and the converging lens has power +5 D. Hence the other lens must have power −7 D, which corresponds to f = −1/7 m ≈ −14.3 cm. Wait: using the stated values, the correct result is −14.3 cm, so none of the listed options is valid.

Explanation:
Using f_eq = 1/P_eq, f_eq = 1/(−2.5) m = −0.40 m. A negative focal length indicates a diverging equivalent lens whose virtual focus lies on the incident side.

Explanation:
The concave lens has power −5 D, while the equivalent power is −2 D. Therefore, the convex lens must contribute +3 D, giving f = 1/3 m = 0.333 m = +33.3 cm.

Explanation:
For identical lenses, P_eq = 2P. Hence each lens has power half the equivalent power: P = (1/−0.25)/2 = −2 D. Therefore, each focal length is −0.50 m = −50 cm.

Explanation:
The equivalent power is −4 D, and the known diverging lens has power −2 D. The second lens must therefore have power −2 D, corresponding to focal length −0.50 m = −50 cm.

Explanation:
The equivalent power is −8 D, while the convex lens has power +4 D. Thus, the diverging lens must have power −12 D, giving f = 1/(−12) m = −0.0833 m = −8.33 cm.