Final answer:
(a) If the pool is completely filled with water (\( d = 2.40 \, \text{m} \)), then the apparent depth (\(d'\)) is calculated as \(d' = \frac{2.40}{1.333}\), resulting in approximately 1.80 m. Consequently, the bottom of the pool seems to be positioned 1.80 m below ground level.
(b) Alternatively, when the pool is half-filled with water (\( d = 2.40/2 = 1.20 \, \text{m} \)), the corresponding apparent depth (\(d'\)) is found using \(d' = \frac{1.20}{1.333}\), yielding about 0.90 m. Thus, in this scenario, the bottom of the pool appears to be approximately 0.90 m below ground level.
Step-by-step explanation:
The apparent depth (\(d'\)) of an object submerged in a transparent medium can be calculated using the formula:
\[ d' = \frac{d}{n} \]
where:
- \( d' \) is the apparent depth,
- \( d \) is the actual depth,
- \( n \) is the refractive index of the medium.
Given that the pool is filled with water, the refractive index (\( n \)) is 1.333.
(a) When the pool is completely filled with water (\( d = 2.40 \, \text{m} \)):
\[ d' = \frac{2.40}{1.333} \approx 1.80 \, \text{m} \]
So, the bottom of the pool appears to be located 1.80 m below ground level.
(b) When the pool is filled halfway with water (\( d = 2.40/2 = 1.20 \, \text{m} \)):
\[ d' = \frac{1.20}{1.333} \approx 0.90 \, \text{m} \]
In this case, the bottom of the pool appears to be located 0.90 m below ground level.