Two electric lamps A and B radiate the same power. Their filaments have the same diemensions, and have emissivities. `e_(A)` and `e_(B)`. Their surface tempratures are `T_(A)` an `T_(B)`. The ratio `T_(A)//T_(B)` will be equal to

Options

A. (a)`((e_(B))/(e_(A)))^(1//4)`
B. (b)`((e_(B))/(e_(A)))^(1//2)`
C. (c)`((e_(A))/(e_(B)))^(1//2)`
D. (d)`((e_(A))/(e_(B)))^(1//4)`

Answer

To solve the problem, we need to find the ratio of the surface temperatures \( T_A \) and \( T_B \) of two electric lamps A and B that radiate the same power.

### Step-by-Step Solution:

1. **Understand the Power Radiated**:
The power radiated by an object due to thermal radiation can be expressed using the Stefan-Boltzmann law:
\[ P = e \cdot A \cdot \sigma \cdot T^4 \]
where \( P \) is the power, \( e \) is the emissivity, \( A \) is the area, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature.

2. **Set Up the Equations for Both Lamps**:
Since both lamps A and B radiate the same power, we can write:
\[ P_A = e_A \cdot A \cdot \sigma \cdot T_A^4 \]
\[ P_B = e_B \cdot A \cdot \sigma \cdot T_B^4 \]
Given that \( P_A = P_B \), we can equate these two expressions.

3. **Equate the Power Expressions**:
\[ e_A \cdot A \cdot \sigma \cdot T_A^4 = e_B \cdot A \cdot \sigma \cdot T_B^4 \]
Since the area \( A \) and the Stefan-Boltzmann constant \( \sigma \) are the same for both lamps, we can cancel them out:
\[ e_A \cdot T_A^4 = e_B \cdot T_B^4 \]

4. **Rearrange to Find the Temperature Ratio**:
We can rearrange the equation to express the ratio of the temperatures:
\[ \frac{T_A^4}{T_B^4} = \frac{e_B}{e_A} \]
Taking the fourth root on both sides gives:
\[ \frac{T_A}{T_B} = \left( \frac{e_B}{e_A} \right)^{1/4} \]

5. **Final Expression**:
Thus, the ratio of the temperatures \( T_A \) and \( T_B \) is:
\[ \frac{T_A}{T_B} = \left( \frac{e_B}{e_A} \right)^{1/4} \]

### Conclusion:
The ratio \( \frac{T_A}{T_B} \) is equal to \( \left( \frac{e_B}{e_A} \right)^{1/4} \).