Question

When an object is placed at a distance of 0.18m in front of a convex lens, a real
image, double in size is produced. Where must the object be placed so that its real
image is three times as big as the object?​

Answer 1

Answer: -0.16m or 0.16m in front of the mirror.

Explanation:

Answer 2

Answer:

Object must be placed at distance 0.24 m in front of the lens for the image to be three times bigger than that of the object.

Explanation:

Sign Convention used:

All distance which are measured along the right side of the lens and above the principal axis of the lens area taken as positive while those which measured along the left side of the lens and below the principal axis are taken as negative.

Given:

Object distance from the lens, $u=-0.18\ m$.

Image formed is double sized, which means, magnification, $m=2$

Let $v$ and $f$ be the image distance from the lens and the focal length of the lens.

Magnification of the lens is given as

$m=\dfrac vu = 2\\v=2u = 2\times -0.18 = -0.36\ m.$

Using lens equation,

$\dfrac 1f=\dfrac 1v-\dfrac 1u\\=\dfrac {1}{-0.36} - \dfrac{1}{-0.18}\\=\dfrac {-1}{0.36}+\dfrac{1}{0.18}\\=\dfrac{-0.18+0.36}{0.18\times 0.36} \\=\dfrac{0.18}{0.18\times 0.36} \\=\dfrac{1}{0.36}\\\Rightarrow f = 0.36\ m.$

Now, for the object to be three times as big as the object, magnification should be 3.

Therefore,

$m=\dfrac vu =3\\v=3u$.

Again using the lens equation for this image,

$\dfrac 1f=\dfrac 1v-\dfrac 1u\\=\dfrac {1}{3u}-\dfrac 1u \\=\dfrac{1-3}{3u}=\dfrac{-2}{3u} \\\Rightarrow u=\dfrac{-2f}3 = \dfrac{-2\times 0.36}{3} = 0.24\ m.$

Thus, at the image will formed three times larger when object is placed at distance 0.24 m in front of the lens.