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(C) When water is poured into the beaker,the apparent depth of the coin changes due to refraction.The apparent depth $d'$ is given by $d' = \frac{h}{\

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$2.5 \, cm$ upwards

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(C) When water is poured into the beaker,the apparent depth of the coin changes due to refraction.The apparent depth $d'$ is given by $d' = \frac{h}{\mu}$,where $h$ is the real depth and $\mu$ is the refractive index of water.The shift in the position of the coin is given by $\Delta d = h - d' = h(1 - \frac{1}{\mu})$.Taking the refractive index of water $\mu \approx \frac{4}{3}$ and real depth $h = 10 \, cm$:$\Delta d = 10 	imes (1 - \frac{1}{4/3}) = 10 	imes (1 - \frac{3}{4}) = 10 	imes \frac{1}{4} = 2.5 \, cm$.Since the coin appears to be raised,the microscope must be moved $2.5 \, cm$ upwards to focus on the new apparent position.

$A$ coin is placed at the bottom of an empty beaker and is focused by a microscope. When water is poured into the beaker to a height of $10 \, cm$,how much must the microscope be moved to focus it again?

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