A real image is obtained at a distance of \(20\) cm from the lens P. When one more lens Q is kept in contact with this lens, the image is shifted \(10\) cm towards the combined lenses. The focal length of the lens Q is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(20\,\mathrm{cm}\)
For thin lenses in contact the powers add: \( \frac{1}{F} = \frac{1}{f_P} + \frac{1}{f_Q} \). The object distance u is the same in both cases, so it cancels.
With lens P alone (image at \( v_1 = +20\,\mathrm{cm} \)):
\( \displaystyle \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_P} \)
With P and Q in contact (image shifted 10 cm towards the lenses, \( v_2 = +10\,\mathrm{cm} \)):
\( \displaystyle \frac{1}{v_2} - \frac{1}{u} = \frac{1}{f_P} + \frac{1}{f_Q} \)
Subtracting the first equation from the second:
\( \displaystyle \frac{1}{f_Q} = \frac{1}{v_2} - \frac{1}{v_1} \)
\( \displaystyle \frac{1}{f_Q} = \frac{1}{10\,\mathrm{cm}} - \frac{1}{20\,\mathrm{cm}} = \frac{1}{20\,\mathrm{cm}} \)
\( \displaystyle f_Q = +20\,\mathrm{cm} \)
The positive sign shows that Q is a converging lens, as expected since it brings the image closer.