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(B) From the lens formula,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad \dots(i)$Here,$f = +10 \,cm$ and $u = -15 \,cm$.Using the lens formul

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$8 \,cm \,s^{-1}$ towards the lens

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(B) From the lens formula,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad \dots(i)$Here,$f = +10 \,cm$ and $u = -15 \,cm$.Using the lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.Thus,$v = +30 \,cm$.Differentiating equation $(i)$ with respect to time $t$:$-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$$\frac{dv}{dt} = \left( \frac{v^2}{u^2} ight) \frac{du}{dt}$Given that the source is moving towards the lens,$\frac{du}{dt} = -2 \,cm \,s^{-1}$ (since $u$ is becoming less negative).$\frac{dv}{dt} = \left( \frac{30}{-15} ight)^2 	imes (-2 \,cm \,s^{-1}) = (2)^2 	imes (-2 \,cm \,s^{-1}) = 4 	imes (-2) = -8 \,cm \,s^{-1}$.The negative sign indicates that the image is moving in the direction of decreasing $v$ (towards the lens). However,since the image is formed on the other side $(v = +30)$,a negative rate of change for $v$ means the image is moving towards the lens. Wait,let's re-evaluate: If $v$ is positive and $\frac{dv}{dt}$ is negative,the distance from the lens is decreasing. Therefore,the image is moving towards the lens at $8 \,cm \,s^{-1}$.

$A$ point source of light is moving at a rate of $2 \,cm \,s^{-1}$ towards a thin convex lens of focal length $10 \,cm$ along its optical axis. When the source is $15 \,cm$ away from the lens,the image is moving at:

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