Answer

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**Hint:**We will start with deducing the lens formula and substituting the focal length and object distance which is provided in the question. By using the lens formula we find the image distance. By using the magnification formula of the lens we will find the size, nature, and position of the image formed.

Formula used

$ \Rightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

**Complete solution:**

Now for the image distance, we will use the lens formula. Lens formula shows the relationship between the image distance $(v)$, object distance $(u)$, and focal length $(f)$.

$ \Rightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ ------------ Equation $(1)$

Now substituting the value of object distance $(v)$, the focal length $(f)$ in the lens formula in the Equation $(1)$

$ \Rightarrow u = 10cm$

$ \Rightarrow f = 15cm$

Now after substitution

$ \Rightarrow \dfrac{1}{v} - \dfrac{1}{{ - 10}} = \dfrac{1}{{15}}$

$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{15}} + \dfrac{1}{{ - 10}}$

$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{15}} - \dfrac{1}{{10}}$

$ \Rightarrow \dfrac{1}{v} = \dfrac{{10 - 15}}{{150}}$

$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 5}}{{150}}$

Now rearranging the values we get,

$ \Rightarrow v = \dfrac{{ - 150}}{5}$

$\therefore v = - 30cm$

Hence the image distance $v = - 30cm$

The negative sign of the image distance $(v)$ shows that the image is on the same side of the lens.

Now by using magnification $(m)$

$ \Rightarrow magnification(m) = \dfrac{{{h_i}}}{h} = \dfrac{v}{u}$

Where ${h_i}$ is the height of the image and the $h$ is object height

$ \Rightarrow \dfrac{{{h_i}}}{h} = \dfrac{v}{u}$

$ \Rightarrow {h_i} = \dfrac{v}{u} \times h$

$ \Rightarrow {h_i} = \dfrac{{ - 30}}{{ - 10}} \times 6$

$\therefore {h_i} = + 18cm$

Now the magnification $(m)$ will be

$ \Rightarrow magnification(m) = \dfrac{v}{u}$

$ \Rightarrow magnification(m) = \dfrac{{ - 30}}{{ - 10}}$

$ \Rightarrow magnification(m) = + 3$

The positive sign of magnification $(m)$and image height $({h_i})$ shows that our image will be virtual and erect. Thus a virtual, erect, and $18cm$ tall image is formed at a distance of ${\text{30}}cm$ on the same side of the lens. The image is three times enlarged as object height is $6cm$ and image height is $18cm$.

**Note**Here the sign of the quantities should be handled carefully. Object distance should always be considered as negative and image distance should be negative if it is formed on the same side where the object is situated while it should be positive when formed on the other side of the lens.

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