Step 1: Understand apparent depth.
When an object is placed at the bottom of a liquid of refractive index $\mu$, the apparent depth is given by $\text{Apparent depth} = \dfrac{\text{Real depth}}{\mu}$.
Step 2: Find the refractive index of water.
For water, real depth $= 12\,\text{cm}$ and apparent depth $= 9\,\text{cm}$. So $\mu_{\text{water}} = \dfrac{12}{9} = \dfrac{4}{3} \approx 1.33$. This is the well-known refractive index of water.
Step 3: Calculate apparent depth in the new liquid.
The real depth of the tank remains the same: $12\,\text{cm}$. For the new liquid with $\mu = 1.5$: \[ \text{Apparent depth} = \frac{12}{1.5} = 8\,\text{cm} \]
Step 4: Find the shift in microscope position.
Initially the microscope is focused at the apparent depth of $9\,\text{cm}$ (in water). After replacing with the new liquid, the apparent depth becomes $8\,\text{cm}$. The needle now appears to be $1\,\text{cm}$ closer to the observer.
Step 5: Determine the direction of movement.
Since the apparent depth decreased from $9\,\text{cm}$ to $8\,\text{cm}$, the image of the needle moves upward (toward the observer). Therefore, the microscope must be moved downward (toward the tank) to refocus on the needle.
Step 6: State the final answer.
The distance through which the microscope has to be moved is: \[ \Delta = 9 - 8 = 1\,\text{cm} \] \[ \boxed{1\,\text{cm}} \]