A screen is placed at a distance of 90 cm from an object. The image of the object on the screen is formed by a convex lens placed at two different locations separated by 20 cm. The focal length of the lens is approximately
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 21.4 cm
In this displacement method the two lens positions are interchangeable: the object distance in one position is the image distance in the other. With object-screen distance \( D \) and lens shift \( d \), the object and image distances are
\( \displaystyle u + v = D = 90\,\mathrm{cm}, \quad v - u = d = 20\,\mathrm{cm} \)
\( v = 55\,\mathrm{cm} \)
\( u = 35\,\mathrm{cm} \)
(distances as magnitudes). For a real image the lens formula gives
\( \displaystyle \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \;\Rightarrow\; f = \frac{uv}{u + v} = \frac{35 \times 55}{90} \)
\( \displaystyle f = \frac{1925}{90} \approx 21.4\,\mathrm{cm} \)
Shortcut: Use the standard result of the displacement method directly:
\( \displaystyle f = \frac{D^{2} - d^{2}}{4D} = \frac{8100 - 400}{4 \times 90} = \frac{7700}{360} \approx 21.4\,\mathrm{cm} \)
Note: A similar question was also asked in CUET (UG) 2022 on 08-08-2022 (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):
Exercises – …
