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(A) The optical device consists of three identical convex lenses with focal length $f = 10 \,cm$,placed at a distance of $30 \,cm$ from each other.Cas

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(A) The optical device consists of three identical convex lenses with focal length $f = 10 \,cm$,placed at a distance of $30 \,cm$ from each other.Case $1$: The source $O$ is at a distance of $10 \,cm$ from the first lens. Since $u = -10 \,cm$ and $f = 10 \,cm$,the rays become parallel after passing through the first lens. These parallel rays fall on the second lens,which focuses them at its focal point,$10 \,cm$ behind it. Since the distance between the second and third lens is $30 \,cm$,the rays from the focal point of the second lens act as a diverging source for the third lens at a distance of $20 \,cm$ $(30 - 10 = 20 \,cm)$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for the third lens,where $u = -20 \,cm$ and $f = 10 \,cm$,we get $\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$,so $v = 20 \,cm$. The final image is formed $20 \,cm$ behind the third lens.Case $2$: When the device is moved away by $10 \,cm$,the source is now at a distance of $20 \,cm$ from the first lens. For the first lens,$u = -20 \,cm$ and $f = 10 \,cm$,so $v = 20 \,cm$. The rays then fall on the second lens at a distance of $10 \,cm$ $(30 - 20 = 10 \,cm)$. Since $u = -10 \,cm$ and $f = 10 \,cm$,the rays become parallel after the second lens. These parallel rays fall on the third lens and are focused at its focal point,$10 \,cm$ behind it.Comparing the final image positions relative to the source,we find that the total distance between the source and the image remains constant in both configurations. Therefore,the shift in the image position is $0 \,cm$.

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