A baby's mouth is `30 cm` from her father's ear and `3.0 m` from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother.

Explanation

To solve the problem of finding the difference in sound intensity levels heard by the father and the mother, we can follow these steps:

### Step 1: Define the distances
Let:
- \( R_f \) = distance from the baby's mouth to the father's ear = 30 cm = 0.3 m
- \( R_m \) = distance from the baby's mouth to the mother's ear = 3.0 m

### Step 2: Understand the relationship between intensity and distance
The intensity of sound (\( I \)) is inversely proportional to the square of the distance (\( R \)). This can be expressed mathematically as:
\[ \frac{I_f}{I_m} = \frac{R_m^2}{R_f^2} \]
Where:
- \( I_f \) = sound intensity at the father's ear
- \( I_m \) = sound intensity at the mother's ear

### Step 3: Calculate the ratio of intensities
Substituting the distances into the equation:
\[ \frac{I_f}{I_m} = \frac{(3.0)^2}{(0.3)^2} = \frac{9}{0.09} = 100 \]

### Step 4: Use the formula for sound intensity levels
The sound intensity level (\( L \)) in decibels (dB) is given by:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
Where \( I_0 \) is the reference intensity. The difference in sound intensity levels between the father and the mother can be expressed as:
\[ L_f - L_m = 10 \log \left( \frac{I_f}{I_0} \right) - 10 \log \left( \frac{I_m}{I_0} \right) \]
This simplifies to:
\[ L_f - L_m = 10 \log \left( \frac{I_f}{I_m} \right) \]

### Step 5: Substitute the ratio of intensities
Using the ratio we calculated in Step 3:
\[ L_f - L_m = 10 \log(100) \]

### Step 6: Calculate the logarithm
Since \( \log(100) = 2 \):
\[ L_f - L_m = 10 \times 2 = 20 \text{ dB} \]

### Final Answer
The difference between the sound intensity levels heard by the father and the mother is **20 dB**.

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