Two thin lenses of powers +15 D and −5 D are placed in contact with each other. An object is placed at 30 cm from this combination. The distance of the final image from the combination is:-
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15 cm
For thin lenses in contact the powers add, and the combination then acts as a single lens in the lens formula. Cartesian sign convention: the object is on the left, so u = −30 cm.
\( P = P_1 + P_2 = +15\,\mathrm{D} - 5\,\mathrm{D} = +10\,\mathrm{D} \)
\( \displaystyle f = \frac{1}{P} = \frac{1}{10\,\mathrm{D}} = 0.1\,\mathrm{m} = +10\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \displaystyle \frac{1}{v} = \frac{1}{10} + \frac{1}{-30} = \frac{3 - 1}{30} = \frac{1}{15} \)
\( v = +15\,\mathrm{cm} \)
v is positive, so a real image is formed 15 cm from the combination, on the side opposite to the object.