A beam of light converges at a point M. Now a concave lens of focal length 32 cm is placed in the path of convergent beam 24 cm from M. Now the beam will converge at:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( +96 \) cm from lens
The beam was converging towards M, which lies beyond the lens, so M acts as a virtual object for the lens. Use the lens formula with the Cartesian sign convention (distances positive in the direction of the incident light).
\( u = +24\,\mathrm{cm}\ (\text{virtual object}) \)
\( f = -32\,\mathrm{cm}\ (\text{concave lens}) \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
\( \displaystyle \frac{1}{v} = -\frac{1}{32} + \frac{1}{24} = \frac{-3 + 4}{96} = \frac{1}{96} \)
\( \displaystyle v = +96\,\mathrm{cm} \)
The positive sign means the beam still converges, now at a real point 96 cm beyond the lens.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
