Answer

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**Hint:**Use lens formula to determine the image distance.

Differentiate the lens formula to determine the formula for rate of movement of image with respect to time.

**Formula used:**

\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]

Here, f is the focal length, v is the image distance and u is the object distance.

**Complete step by step answer:**

We know that the focal length of the convex lens is positive. Also, the object distance from the lens is taken as negative.

We have the lens formula,

\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]

Here, f is the focal length, v is the image distance and u is the object distance.

Rearrange the above equation to determine the image distance as follows,

\[v = \dfrac{{fu}}{{u + f}}\]

Substitute 10 cm for f and \[ - 15\,cm\] for u in the above equation.

\[v = \dfrac{{\left( {10\,cm} \right)\left( { - 15\,cm} \right)}}{{ - 15\,cm + 10\,cm}}\]

Differentiate the lens formula with respect to time t as follows,

\[0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]

\[ \Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]

\[\therefore \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}\]

Substitute \[ + 30\,cm\] for v, \[ - 15\,cm\] for u and \[ + 2\,cm - {s^{ - 2}}\] for \[\dfrac{{du}}{{dt}}\] in the above equation.

\[\dfrac{{dv}}{{dt}} = \dfrac{{{{\left( {30\,cm} \right)}^2}}}{{{{\left( { - 15\,cm} \right)}^2}}}\left( {2\,cm - {s^{ - 2}}} \right)\]

\[\dfrac{{dv}}{{dt}} = \left( 4 \right)\left( {2\,cm - {s^{ - 2}}} \right)\]

\[\dfrac{{dv}}{{dt}} = + 8\,cm - {s^{ - 2}}\]

The positive sign in the above equation implies that the image is moving along the positive direction of the x-axis which is away from the lens

**So, the correct answer is “Option D”.**

**Note:**

Read the properties of the image formed by the convex lens. The focal length of the convex lens is positive whereas the focal length of the concave lens is negative. The object distance or source distance is negative while the image distance is positive.

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