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.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

318 Views

Kickstart Your Career

Get certified by completing the course

Get Started