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. A) 16P
B. B) 4P
C. C) 2P
D. D) P

Answer

To solve the problem, we need to apply the Stefan-Boltzmann law and the inverse square law of radiation. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
We have a small sphere at temperature \( T \) radiating power \( P \) to a black metal foil located at a distance \( d \).

### Step 2: Apply the Stefan-Boltzmann Law
According to the Stefan-Boltzmann law, the power received by the black metal foil (which behaves as a perfect black body) is proportional to the fourth power of the absolute temperature of the source.

The power received by the foil can be expressed as:
\[ P \propto T^4 \]

### Step 3: Calculate the New Temperature
If the temperature \( T \) is doubled, the new temperature \( T_2 \) becomes:
\[ T_2 = 2T \]

### Step 4: Calculate the New Power Received
Using the Stefan-Boltzmann law again, the new power \( P_2 \) received by the foil when the temperature is doubled is:
\[ P_2 = k (T_2)^4 = k (2T)^4 = k \cdot 16T^4 \]
Thus, the new power received by the foil is:
\[ P_2 = 16P \]

### Step 5: Consider the Effect of Distance
Now, we also need to consider the effect of distance. If the distance \( d \) is doubled, the new distance \( d_2 \) becomes:
\[ d_2 = 2d \]

According to the inverse square law, the intensity \( I \) of radiation at a distance \( d \) from a point source is given by:
\[ I \propto \frac{P}{d^2} \]

### Step 6: Calculate the New Intensity
The new intensity \( I_2 \) at distance \( d_2 \) is:
\[ I_2 = \frac{P_2}{(d_2)^2} = \frac{16P}{(2d)^2} = \frac{16P}{4d^2} = \frac{4P}{d^2} \]

### Step 7: Final Power Received by the Foil
The power received by the foil is proportional to the intensity and the area of the foil. Since the area remains constant, the new power \( P_2 \) received by the foil is:
\[ P_2 = I_2 \cdot A = \left(\frac{4P}{d^2}\right) \cdot A \]

### Step 8: Conclusion
Since the area of the foil does not change, we conclude that the power received by the foil when both the temperature and the distance are doubled is:
\[ P_2 = 4P \]

### Final Answer
Thus, the power received by the foil will be \( 4P \).

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