The two possible positions of an object kept in front of a lens of power \( +10\ \mathrm{D} \), so that the image formed in both cases is five times magnified, are:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( u_{1} = -12\ \mathrm{cm},\ u_{2} = -8\ \mathrm{cm} \)
A convex lens gives a magnified image in two ways: a real, inverted image (m = −5) when the object is between f and 2f, and a virtual, erect image (m = +5) when it is inside f. Sign convention: distances measured from the lens, positive in the direction of the incident light, so a real object has u negative.
\( f = 1/P = 1/(10\ \mathrm{D}) = 0.1\ \mathrm{m} = +10\ \mathrm{cm} \)
\( m = v/u \)
Lens formula:
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
Real image, m = −5, so v = −5u:
\( \displaystyle -\frac{1}{5u} - \frac{1}{u} = \frac{1}{10} \quad\Rightarrow\quad -\frac{6}{5u} = \frac{1}{10} \quad\Rightarrow\quad u_{1} = -12\ \mathrm{cm} \)
Virtual image, m = +5, so v = +5u:
\( \displaystyle \frac{1}{5u} - \frac{1}{u} = \frac{1}{10} \quad\Rightarrow\quad -\frac{4}{5u} = \frac{1}{10} \quad\Rightarrow\quad u_{2} = -8\ \mathrm{cm} \)
Shortcut: Eliminating v from the lens formula gives \( m = f/(f + u) \), so \( u = f(1/m - 1) \): m = −5 gives u = 10(−1/5 − 1) = −12 cm, and m = +5 gives u = 10(1/5 − 1) = −8 cm.