Two thin lenses of focal length \( f_{1} \) and \( f_{2} \) are placed co-axially in contact. The power of the combination is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{f_{1}+f_{2}}{f_{1}f_{2}} \)
The power of a lens is the reciprocal of its focal length, \( P = 1/f \), and for thin lenses in contact the powers add:
\( \displaystyle P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} \)
\( \displaystyle P = \frac{f_1 + f_2}{f_1 f_2} \)
Shortcut: Power has the unit \( \mathrm{m^{-1}} \), while Options (1) and (2) are pure numbers, so they are out. For two identical lenses (\( f_1 = f_2 = f \)) the power must be \( 2/f \): Option (3) gives zero and Option (4) gives \( 2/f \).
Note: A similar question was also asked in CUET (UG) 2023 on 13-06-2023 (Slot 1) and on 12-06-2023 (Slot 3).
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.”