What would be the position of the image formed by the lens combinatio…

The correct answer is Option (2) → 30 cm to the right of the third lens

For lenses in series, the image formed by one lens is the object for the next. Use the lens formula with the Cartesian sign convention (distances measured from each lens, positive in the direction of the incident light), with all distances in cm.

    \( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)

First lens: \( u_1 = -30\,\mathrm{cm} \), \( f_1 = +10\,\mathrm{cm} \)

    \( \displaystyle \frac{1}{v_1} = \frac{1}{10} + \frac{1}{-30} = \frac{3 - 1}{30} = \frac{1}{15} \)

    \( \displaystyle v_1 = +15\,\mathrm{cm} \)

The second lens is 5 cm to the right of the first, so this image lies 10 cm beyond the second lens; for the second lens it is a virtual object: \( u_2 = +10\,\mathrm{cm} \), \( f_2 = -10\,\mathrm{cm} \)

    \( \displaystyle \frac{1}{v_2} = \frac{1}{-10} + \frac{1}{10} = 0 \)

    \( \displaystyle v_2 = \infty \)

The light leaving the concave lens is a parallel beam. A parallel beam falling on the third lens (\( f_3 = +30\,\mathrm{cm} \)) converges at its focus, so the final image is 30 cm to the right of the third lens.

Note: It seems that there is a mistake in the original NTA paper, which prints the focal lengths of the first two lenses in the figure as +10 and −10 without a unit when they should read +10 cm and −10 cm. All other distances in the figure are in cm, so the question can still be solved.

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 – …