A 20 mm thick layer of water (`n=1.33`) floats on a 40 mm thick layer of carbon tetrachloride `(n=1.46)` in a tank. A coin lies at the bottom of the tank. At what depth below the top water surface do you perceive the coin ?

Answer

To solve the problem of determining the perceived depth of the coin lying at the bottom of a tank with two layers of different liquids, we will use the concept of apparent depth and refractive indices.

### Step-by-step Solution:

1. **Identify the Layers and Their Properties**:
- The top layer is water with a thickness of \( h_1 = 20 \, \text{mm} \) and a refractive index \( n_1 = 1.33 \).
- The bottom layer is carbon tetrachloride (CCl4) with a thickness of \( h_2 = 40 \, \text{mm} \) and a refractive index \( n_2 = 1.46 \).

2. **Calculate the Apparent Depth in Each Medium**:
- The formula for apparent depth \( h_a \) in a medium is given by:
\[ h_a = \frac{h_r}{n} \]
where \( h_r \) is the real depth and \( n \) is the refractive index of the medium.

3. **Calculate Apparent Depth for Water**:
- For the water layer:
\[ h_{a1} = \frac{h_1}{n_1} = \frac{20 \, \text{mm}}{1.33} \approx 15.04 \, \text{mm} \]

4. **Calculate Apparent Depth for Carbon Tetrachloride**:
- For the carbon tetrachloride layer:
\[ h_{a2} = \frac{h_2}{n_2} = \frac{40 \, \text{mm}}{1.46} \approx 27.40 \, \text{mm} \]

5. **Total Apparent Depth**:
- The total apparent depth \( h_a \) perceived from the top of the water surface is the sum of the apparent depths from both layers:
\[ h_a = h_{a1} + h_{a2} \approx 15.04 \, \text{mm} + 27.40 \, \text{mm} \approx 42.44 \, \text{mm} \]

6. **Final Result**:
- Therefore, the perceived depth of the coin below the top water surface is approximately \( 42.4 \, \text{mm} \).