The power of the composite lens made of two thin lenses of focal lengths \( f_1 \) and \( f_2 \) placed in contact will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{(f_1+f_2)}{f_1f_2} \)
The power of a lens is the reciprocal of its focal length, \( P = 1/f \), and for thin lenses in contact the powers add.
\( P = P_1 + P_2 \)
\( \displaystyle P = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_2 + f_1}{f_1f_2} \)
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1), and in CUET (UG) 2023 on 13-06-2023 (Slot 1).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Combination of Thin Lenses in Contact – “… If several thin lenses of focal length f1, f2, f3,... are in contact, the effective focal length of their combination is given by
\( \displaystyle \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \ldots \qquad (9.31) \)
In terms of power, Eq. (9.31) can be written as
\( \displaystyle P = P_1 + P_2 + P_3 + \ldots \qquad (9.32) \)
where P is the net power of the lens combination.”