A black metal foil is warmed by radiation from a small sphere at temperature `T` and at a distance d it is found that the power received by the foil is `P` If both the temperature and the distance are doubled the power received by the foil will be .

Options

A. `64P`
B. `16P`
C. `4P`
D. `8P`

Explanation

To solve the problem, we will use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and inversely proportional to the square of the distance from the source.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the initial temperature of the sphere be \( T_1 \).
- The initial distance from the sphere to the foil is \( d_1 \).
- The power received by the foil at this distance is \( P \).

2. **Applying the Stefan-Boltzmann Law**:
- The power \( P \) received by the foil can be expressed as:
\[ P = \sigma A \left( \frac{T_1^4}{d_1^2} \right) \]
- Here, \( \sigma \) is the Stefan-Boltzmann constant, and \( A \) is the area of the foil.

3. **Changing the Conditions**:
- According to the problem, both the temperature and the distance are doubled:
- New temperature \( T_2 = 2T_1 \)
- New distance \( d_2 = 2d_1 \)

4. **Calculating the New Power**:
- Using the new conditions, the power \( P' \) received by the foil can be expressed as:
\[ P' = \sigma A \left( \frac{T_2^4}{d_2^2} \right) \]
- Substituting \( T_2 \) and \( d_2 \):
\[ P' = \sigma A \left( \frac{(2T_1)^4}{(2d_1)^2} \right) \]
- Simplifying this expression:
\[ P' = \sigma A \left( \frac{16T_1^4}{4d_1^2} \right) = \sigma A \left( \frac{16}{4} \cdot \frac{T_1^4}{d_1^2} \right) = 4 \sigma A \left( \frac{T_1^4}{d_1^2} \right) \]
- Thus, we find that:
\[ P' = 4P \]

5. **Conclusion**:
- Therefore, when both the temperature and the distance are doubled, the power received by the foil will be \( 4P \).

### Final Answer:
The power received by the foil when both the temperature and distance are doubled is \( 4P \).