In Y.D.S.E., slits are at a distance of 0.5 mm and light of wavelength `lambda` is used. Determine the value of `lambda` if interference pattern is formed at a distance 1.2 m away and third dark fringe is at a distance of 1.2 cm from the central fringe.

Answer

To solve the problem, we will use the formula for the position of the nth dark fringe in a Young's Double Slit Experiment (YDSE). The formula for the position of the nth dark fringe is given by:

\[ Y_n = \frac{(n - 1) \lambda D}{d} \]

Where:
- \( Y_n \) = position of the nth dark fringe from the central maximum
- \( n \) = order of the dark fringe (for the third dark fringe, \( n = 3 \))
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Given:
- Distance between the slits, \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1.2 \, \text{m} \)
- Position of the third dark fringe, \( Y_3 = 1.2 \, \text{cm} = 1.2 \times 10^{-2} \, \text{m} \)

### Step-by-step Solution:

1. **Identify the order of the dark fringe**:
For the third dark fringe, \( n = 3 \).

2. **Substitute the values into the formula**:
\[ Y_3 = \frac{(3 - 1) \lambda D}{d} \]
This simplifies to:
\[ Y_3 = \frac{2 \lambda D}{d} \]

3. **Rearranging the equation to solve for \( \lambda \)**:
\[ \lambda = \frac{Y_3 \cdot d}{2D} \]

4. **Substituting the known values**:
\[ \lambda = \frac{(1.2 \times 10^{-2} \, \text{m}) \cdot (0.5 \times 10^{-3} \, \text{m})}{2 \cdot (1.2 \, \text{m})} \]

5. **Calculating the numerator**:
\[ 1.2 \times 10^{-2} \cdot 0.5 \times 10^{-3} = 0.6 \times 10^{-5} \, \text{m}^2 \]

6. **Calculating the denominator**:
\[ 2 \cdot 1.2 = 2.4 \, \text{m} \]

7. **Now substituting these values back into the equation for \( \lambda \)**:
\[ \lambda = \frac{0.6 \times 10^{-5}}{2.4} \]

8. **Perform the division**:
\[ \lambda = 0.25 \times 10^{-5} \, \text{m} \]

9. **Convert to standard form**:
\[ \lambda = 2.5 \times 10^{-6} \, \text{m} \]

### Final Answer:
The wavelength \( \lambda \) is \( 2.5 \, \mu m \) or \( 2.5 \times 10^{-6} \, \text{m} \).