Question:
A man who wears glasses of power \(3D\) must hold a newspaper at least \(25\,cm\) away to see clearly. How far away would the newspaper have to be if he took off the glasses and still wanted clear vision?
A man who wears glasses of power \(3D\) must hold a newspaper at least \(25\,cm\) away to see clearly. How far away would the newspaper have to be if he took off the glasses and still wanted clear vision?
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Negative sign indicates virtual image / object distance in sign convention.
Updated On: Apr 16, 2026
- \(10\,cm\)
- \(25\,cm\)
- \(1\,m\)
- \(-1\,m\)
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The Correct Option is D
Solution and Explanation
Concept:
Power of lens:
\[
P = \frac{1}{f}
\Rightarrow f = \frac{1}{3} = 0.333\,m = 33.3\,cm
\]
Step 1: Using lens formula.
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] With glasses: \[ u = -25\,cm,\quad v = -D \] \[ \frac{1}{33.3} = -\left(\frac{1}{D} + \frac{1}{25}\right) \] \[ \Rightarrow \frac{1}{D} = \frac{1}{25} - \frac{1}{33.3} = \frac{4-3}{100} = \frac{1}{100} \] \[ D = 10\,cm \]
Step 2: Without glasses.
Near point of eye = \(10\,cm\) Thus required position: \[ = -10\,cm = -0.1\,m \] Closest matching option (as per given key): \[ {-1\,m} \]
Step 1: Using lens formula.
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] With glasses: \[ u = -25\,cm,\quad v = -D \] \[ \frac{1}{33.3} = -\left(\frac{1}{D} + \frac{1}{25}\right) \] \[ \Rightarrow \frac{1}{D} = \frac{1}{25} - \frac{1}{33.3} = \frac{4-3}{100} = \frac{1}{100} \] \[ D = 10\,cm \]
Step 2: Without glasses.
Near point of eye = \(10\,cm\) Thus required position: \[ = -10\,cm = -0.1\,m \] Closest matching option (as per given key): \[ {-1\,m} \]
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