In a Young.s interference experimental arrangement, the yellow light is composed of two wavelengths `5890 A^(0)" and "5895 A^(0)`. The distance between the two slits is `10^(-3)m` and screen is placed 1m away. Upto what order can fringes be seen?

Options

A. `589`
B. `280`
C. `440`
D. `339`

Answer

To solve the problem of determining the maximum order of fringes that can be seen in a Young's interference experiment with two wavelengths of yellow light, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Wavelength 1, \( \lambda_1 = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \)
- Wavelength 2, \( \lambda_2 = 5895 \, \text{Å} = 5895 \times 10^{-10} \, \text{m} \)
- Distance between the slits, \( d = 10^{-3} \, \text{m} \)
- Distance to the screen, \( L = 1 \, \text{m} \)

2. **Understand the condition for overlapping fringes**:
- The bright fringe of one wavelength overlaps with the dark fringe of the other wavelength. This occurs when:
\[ n \lambda_2 = (n + \frac{1}{2}) \lambda_1 \]

3. **Rearranging the equation**:
- Rearranging the above equation gives:
\[ n \lambda_2 = n \lambda_1 + \frac{1}{2} \lambda_1 \]
- This can be rewritten as:
\[ n (\lambda_2 - \lambda_1) = \frac{1}{2} \lambda_1 \]

4. **Substituting the values**:
- Calculate \( \lambda_2 - \lambda_1 \):
\[ \lambda_2 - \lambda_1 = 5895 \, \text{Å} - 5890 \, \text{Å} = 5 \, \text{Å} = 5 \times 10^{-10} \, \text{m} \]
- Now substitute this into the rearranged equation:
\[ n (5 \times 10^{-10}) = \frac{1}{2} (5890 \times 10^{-10}) \]

5. **Solving for \( n \)**:
- Simplifying gives:
\[ n = \frac{\frac{1}{2} (5890 \times 10^{-10})}{5 \times 10^{-10}} = \frac{5890}{10} = 589 \]

6. **Conclusion**:
- The maximum order of fringes that can be seen is \( n = 589 \).

### Final Answer:
The maximum order of fringes that can be seen is **589**.

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