A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.

Focal length, $f$ = $+$5 cm

Image distance, $v$ = $-$25 cm     (negative sign shows that the image is virtual or placed on the left side of the lens)

To find: Magnification $(m)$.

Solution:

According to the lens formula, we know that:

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values in the formula we get-

$\frac {1}{(-25)}-\frac {1}{u}=\frac {1}{5}$

$-\frac {1}{25}-\frac {1}{u}=\frac {1}{5}$

$-\frac {1}{5}-\frac {1}{25}=\frac {1}{u}$

$\frac {1}{u}=\frac {-5-1}{25}$

$\frac {1}{u}=-\frac {6}{25}$

$u=-\frac {25}{6}$

Thus, the object $u$ is formed at a distance of $\frac {25}{6}$ from the convex lens, and the negative $(-)$ sign for object distance implies that the object is placed on the left side of the convex lens.

Now,

According to magnification formula, we know that:

$m=\frac {v}{u}$

$m=\frac {-25}{-\frac {25}{6}}$

$m=\frac {-25\times {6}}{-25}$

$m=+6$

Thus, the value of the magnification is 6.

As, the magnification of the image is 6, which is more than 1, so the image will be larger than the object. And, the positive sign $(+)$ for magnification implies that the image is formed above the principal axis.