Question:
A card divided into squares each of size \(1\ \mathrm{mm^2}\) is viewed through a magnifying glass of focal length \(10\) cm which is held close to the eye. The distance at which the lens is to be placed from the card to view the squares with maximum possible magnification is
A card divided into squares each of size \(1\ \mathrm{mm^2}\) is viewed through a magnifying glass of focal length \(10\) cm which is held close to the eye. The distance at which the lens is to be placed from the card to view the squares with maximum possible magnification is
Show Hint
A simple microscope produces maximum magnification when the image is formed at the least distance of distinct vision.
Use
\[
\boxed{
v=-25\text{ cm}
}
\]
and apply
\[
\boxed{
\frac1f=\frac1v-\frac1u.
}
\]
Updated On: Jul 18, 2026
- \(20\) cm
- \(16.67\) cm
- \(10\) cm
- \(7.14\) cm
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Recall the condition for maximum magnification.
A magnifying glass gives maximum angular magnification when the final image is formed at the least distance of distinct vision.
Thus,
\[
v=-D=-25\text{ cm}.
\]
The focal length is
\[
f=10\text{ cm}.
\]
Step 2: Apply the lens formula. Using \[ \frac1f=\frac1v-\frac1u, \] we get \[ \frac1{10} = -\frac1{25} -\frac1u. \] Therefore, \[ -\frac1u = \frac1{10}+\frac1{25} = \frac7{50}. \] Hence, \[ u = -\frac{50}{7}\text{ cm}. \]
Step 3: Find the required distance. The distance between the lens and the card is \[ |u| = \frac{50}{7} = 7.14\text{ cm}. \] Hence, \[ \boxed{7.14\text{ cm}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.
Step 2: Apply the lens formula. Using \[ \frac1f=\frac1v-\frac1u, \] we get \[ \frac1{10} = -\frac1{25} -\frac1u. \] Therefore, \[ -\frac1u = \frac1{10}+\frac1{25} = \frac7{50}. \] Hence, \[ u = -\frac{50}{7}\text{ cm}. \]
Step 3: Find the required distance. The distance between the lens and the card is \[ |u| = \frac{50}{7} = 7.14\text{ cm}. \] Hence, \[ \boxed{7.14\text{ cm}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.
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