The end correction ( e) is `(l_(1)` = length of air column at first resonance and `l_(2)` is length of air column at second resonance).

Options

A. `(a)e = (l_2-3l_1)/2`
B. `(b)e = (l_1-3l_2)/2`
C. `(c)e = (l_2-2l_1)/2`
D. `(d)e = (l_1-2l_2)/2`

Explanation

To find the end correction (e) in the context of resonance in a tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Resonance Conditions**:
- At the first resonance, the length of the air column (L1) plus the end correction (e) is equal to one-fourth of the wavelength (λ). This can be expressed as:
\[ L1 + e = \frac{\lambda}{4} \quad \text{(Equation 1)} \]

- At the second resonance, the length of the air column (L2) plus the end correction (e) is equal to three-fourths of the wavelength (λ). This can be expressed as:
\[ L2 + e = \frac{3\lambda}{4} \quad \text{(Equation 2)} \]

2. **Relate the Two Equations**:
- From Equation 1, we can express λ in terms of L1 and e:
\[ \lambda = 4(L1 + e) \]
- Substitute this expression for λ into Equation 2:
\[ L2 + e = \frac{3}{4} \cdot 4(L1 + e) \]
- Simplifying this gives:
\[ L2 + e = 3(L1 + e) \]

3. **Expand and Rearrange**:
- Expanding the right side:
\[ L2 + e = 3L1 + 3e \]
- Rearranging the equation to isolate e:
\[ L2 - 3L1 = 3e - e \]
\[ L2 - 3L1 = 2e \]

4. **Solve for End Correction (e)**:
- Dividing both sides by 2 gives:
\[ e = \frac{L2 - 3L1}{2} \]

5. **Final Expression**:
- Thus, the end correction (e) can be expressed as:
\[ e = \frac{L2 - 3L1}{2} \]

### Conclusion:
The final expression for the end correction (e) is:
\[ e = \frac{L2 - 3L1}{2} \]