Two thin lenses of powers $+5\,\text{D}$ and $-3\,\text{D}$ are placed in CONTACT. The total power of the combination is

1. Your syllabus: when thin lenses are placed in CONTACT, their powers ADD algebraically: $P_\text{total} = P_1 + P_2 + \ldots$. 2. Equivalently, $1/F = 1/f_1 + 1/f_2 + \ldots$ for the focal lengths. 3. Apply to this problem: $P_\text{total} = (+5) + (-3) = +2\,\text{D}$. 4. So the combination is equivalent to a single converging lens of focal length $f = 1/P = 1/2 = 0.5\,\text{m} = 50\,\text{cm}$. 5. Option B multiplies the powers. Option C ignores the sign of the second lens. Option D inverts the sum (reciprocal slip). _Source: syllabus Physics Part 2, Ch 9, §9.6.4 (Combination of thin lenses in contact, Eq. 9.22), p. 18–19._