In YDSE with a light of wavelength 6000 A. Distance between coherent sources and screen is 1.0 m and fringe width is 4 mm. The separation (in mm) between the two coherent sources is

Explanation

To solve the problem, we need to find the separation between the two coherent sources in a Young's Double Slit Experiment (YDSE). We are given the following information:

- Wavelength of light, \( \lambda = 6000 \) Å (angstroms)
- Distance between the coherent sources and the screen, \( D = 1.0 \) m
- Fringe width, \( \beta = 4 \) mm

### Step-by-Step Solution:

1. **Convert Wavelength to Meters:**
The wavelength is given in angstroms. We need to convert it to meters for consistency in units.
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

2. **Convert Fringe Width to Meters:**
The fringe width is given in mm. We will convert it to meters.
\[ \beta = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \]

3. **Use the Formula for Fringe Width:**
The formula for fringe width in YDSE is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \( d \) is the separation between the two coherent sources.

4. **Rearranging the Formula:**
We need to rearrange the formula to solve for \( d \):
\[ d = \frac{\lambda D}{\beta} \]

5. **Substituting the Values:**
Now, we can substitute the values we have:
\[ d = \frac{(6 \times 10^{-7} \, \text{m})(1.0 \, \text{m})}{4 \times 10^{-3} \, \text{m}} \]

6. **Calculating \( d \):**
Performing the calculation:
\[ d = \frac{6 \times 10^{-7}}{4 \times 10^{-3}} = \frac{6}{4} \times 10^{-4} = 1.5 \times 10^{-4} \, \text{m} \]

7. **Convert \( d \) to mm:**
Since we need the answer in mm, we convert meters to mm:
\[ d = 1.5 \times 10^{-4} \, \text{m} = 0.15 \, \text{mm} \]

### Final Answer:
The separation between the two coherent sources is \( 0.15 \, \text{mm} \).