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**Hint:**We all know that a lens is composed of two identical spherical surfaces having the same refractive index different from the surrounding medium. We will have to first apply the formula for refraction at the interface to find out the image distance formed by the first interface. Then that image will act as a virtual object for the second surface. The Lens Maker formula gives this.

**Complete step by step answer:**

We will now write the given data according to sign convention:

The distance of the image is v = -10 cm.

The radius of curvature of the refracting surface is R=10 cm.

We know that the lens maker formula is given by:

$\dfrac{{{n_2}}}{v} - \dfrac{{{n_1}}}{u} = \dfrac{{{n_2} - {n_1}}}{R}$

Here ${n_2} = 1$ is the refractive index of air and ${n_1} = 1.5$ is the refractive index of glass.

We will now substitute the known values in the above equation.

$\begin{array}{l}

\Rightarrow \dfrac{1}{{ - 10\;cm}} - \dfrac{{1.5}}{u} = \dfrac{{1 - 1.5}}{{10\;cm}} \Rightarrow u = 30\;cm

\end{array}$

**Therefore, the actual location of air bubbles is at 30 cm, and the correct option is (C).**

**Additional Information:**If a convex lens, having focal length f, is cut along the principal axis, both the resulting pieces will have the same focal length f. Water droplets can be considered as convex lenses, and the lens maker formula is applicable. The reciprocal of the focal length is called the refractive power that has units dioptre (inverse meter).

**Note:**Sign convention for the distances is very important and should be the same while application of formula on both the interfaces. Drawing separate diagrams and separately applying the formula on both surfaces will reduce the chances of confusion a lot.

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