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**Hint**Calculate the resultant power of the combination of lenses by simple addition. The inverse of this power gives the focal length of this combination. Substituting the value of the resultant value of focal length into the lens formula gives the position of the final image. Magnification can be calculated by dividing the negative of the image distance with the object distance. Equating this value of magnification to the equation of magnification; that is the ratio of image size to the object size, gives the value for image size.

**Complete step by step answer:**

Let the power of the first lens be ${{P}_{1}}$ and its focal length be ${{f}_{1}}$. Let the power of the second lens and its focal length be ${{P}_{2}}$ and ${{f}_{2}}$ respectively. Let the combination of lenses have $P$ and $F$ as the resultant power and resultant focal length of the combination respectively.

Given, ${{P}_{1}}=+15D$ and ${{P}_{2}}=-5D$.

Resultant power of the combination,

$P={{P}_{1}}+{{P}_{2}}$

Substituting the given values of ${{P}_{1}}$ and ${{P}_{2}}$ gives

$P=15-5$

$\Rightarrow P=+10D$

The focal length of a lens is the inverse of its power.

$\therefore F=\dfrac{1}{P}$

Substituting the value of the power of the combination, we get

$F=\dfrac{1}{10}m$

Or

$F=0.1m$

Let the object distance from the lens be $u$, image distance be $v$, magnification be $m$, object size be ${{h}_{1}}$, and image size be $h{}_{2}$ .

By using the lens formula for focal length, we get

$\dfrac{1}{F}=\dfrac{1}{u}+\dfrac{1}{v}$

Rearranging, we get

$\dfrac{1}{v}=\dfrac{1}{F}-\dfrac{1}{u}$

Substituting the given value of object distance, $u=30cm$ and the calculated value of focal length, we get

$\dfrac{1}{v}=10-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{299}{30}$

$\Rightarrow v=\dfrac{30}{299}cm$

Or

$v=0.10034cm$

Magnification can be found using image distance and object distance as follows,

$m=\dfrac{-v}{u}$

Substituting the calculated and given values into the above equation gives us

$m=\dfrac{-30}{299}\times \dfrac{1}{30}$

$\Rightarrow m=\dfrac{-1}{299}$

or

m=$-3.3445\times {{10}^{-3}}$

The general equation for magnification of a lens is given by

$m=\dfrac{{{h}_{2}}}{{{h}_{1}}}$

Substituting the calculated value of $m$ and the given value of object size ${h_1} = 3cm$gives us

$\dfrac{-1}{299}=\dfrac{{{h}_{2}}}{3}$

By rearranging,

$\Rightarrow {{h}_{2}}=$ ${{h}_{2}}=\dfrac{-3}{299}$

$\Rightarrow {{h}_{2}}=-0.010034cm$

**Therefore, the image formed is diminished with a size of $0.010034cm$. The negative sign indicates that the image is inverted.**

**Note**

We used the classical sign convention to approach this problem- all real object and image distances are taken to be positive and all virtual object and image distances are taken to be negative. The focal length of a convex lens is taken to be positive and the focal length of a concave lens is taken to be negative. In some books, you may find that they have used a different sign convention. As long as a single type of sign convention is used throughout a problem, there won’t be an issue.

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