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Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect (upright) image of linear magnification 4.
Given:
Converging lens is a convex lens.
Focal length, $f$ = 10 cm
Magnification, $m$ = $+$4 (positive sign, implies that the image is virtual and erect)
To find: Object distance, $u$ from the convex lens.
Solution:
From the magnification formula, we know that-
$m=\frac {v}{u}$
Substituting the given values in the formula we get-
$4=\frac {v}{u}$
$v=4u$
Now,
From the lens formula, we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the value of $f$ and $v$ in the formula we get-
$\frac {1}{4u}-\frac {1}{u}=\frac {1}{10}$
$\frac {1}{10}=\frac {1-4}{4u}$
$\frac {1}{10}=\frac {-3}{4u}$
$4u=10\times {(-3)}$
$u=\frac {-30}{4}$
$u=-7.5cm$
Thus, the object $u$ must be placed at a distance of 7.5 cm in front of the lens.
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