In Y.D.S.E., light of wavelength `5500 Å` is used. If the slits are at a distance of `10^(-3) m` and interference pattern is formed at a distance of 100 cm.
Calculate the distance between central fringe and second dark fringe.

Answer

To solve the problem of finding the distance between the central fringe and the second dark fringe in Young's Double Slit Experiment (Y.D.S.E.), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of light, \( \lambda = 5500 \, \text{Å} = 5500 \times 10^{-10} \, \text{m} \)
- Distance between the slits, \( d = 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 100 \, \text{cm} = 1 \, \text{m} \)

2. **Formula for the Distance of the nth Dark Fringe:**
The distance of the nth dark fringe from the central maximum is given by the formula:
\[ y_n = \frac{(n - 0.5) \lambda D}{d} \]
For the second dark fringe, \( n = 2 \).

3. **Substituting the Values:**
Substitute \( n = 2 \), \( \lambda = 5500 \times 10^{-10} \, \text{m} \), \( D = 1 \, \text{m} \), and \( d = 10^{-3} \, \text{m} \) into the formula:
\[ y_2 = \frac{(2 - 0.5) \cdot (5500 \times 10^{-10}) \cdot 1}{10^{-3}} \]
Simplifying this gives:
\[ y_2 = \frac{(1.5) \cdot (5500 \times 10^{-10})}{10^{-3}} \]

4. **Calculating the Value:**
\[ y_2 = 1.5 \cdot 5500 \cdot 10^{-10} \cdot 10^{3} \]
\[ y_2 = 1.5 \cdot 5500 \cdot 10^{-7} \]
\[ y_2 = 8250 \times 10^{-7} \, \text{m} = 82.5 \times 10^{-6} \, \text{m} \]

5. **Convert to Millimeters:**
\[ y_2 = 0.825 \, \text{mm} \]

### Final Answer:
The distance between the central fringe and the second dark fringe is \( 0.825 \, \text{mm} \).

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