A single converging lens is held close to the eye for use as a mangnifying glass. For maximum magnifying power, the lens must be positioned, so that
A
the image is at infinity.
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B
object is just beyond the principal focus of the lens.
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C
the image is at the principal focus of the lens.
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D
the image is at the near point of the eye.
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Solution
The correct option is D the image is at the near point of the eye.
Magnifying power
= Angle subtended by the image on the eye when lens is usedAngle subtended on naked eye when object is placed at near point
Suppose object is placed at distance u0from lens, So that its image is formed at near point.
1v−1u=1f
⇒1−D−1−u0=1f
1u0=1D+1f ⇒Du0=1+Df
Angle subtended by image on lens (and hence, on eye) is