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# An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Object distance, $u$ = $-$10 cm (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = $+$20 cm

Object height, $h$ = 4 cm

To find: Position and nature of the image $(v)$, Size of the image $(h')$.

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-

$\frac {1}{v}-\frac {1}{(-10)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{10}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{10}$

$\frac {1}{v}=\frac {1-2}{20}$

$\frac {1}{v}=-\frac {1}{20}$

$v=-20cm$

Thus, the image $v$ is formed at a distance of 20 cm from the convex lens, and the negative $(-)$ sign for image distance implies that the image is formed **on the left side of the convex lens.**

As the image is formed on the left side of the lens, the nature of the image will be **virtual and erect**.

Now,

According to magnification formula, we know that:

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

$m=\frac {-20}{-10}$

$m=+2$

Thus, the magnification of the image is **2, **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.**

Again,

According to the magnification formula, we know that:

$m=\frac {h'}{h}$

$2=\frac {h'}{4}$

$h'=4\times {2}$

$h'=8cm$

Thus, the height or size of the image, $h'$ is **8cm.**

Hence, the position of the image is on the left side of the lens, the nature of the image is virtual and erect, and the **size of the image is** 8cm.