In a biprism experiement, by using light of wavelength `5000Å`, `5mm` wide fringes are obtained on a screen `1.0m` away from the coherent sources. The separation between the two coherent sources is

Options

A. 1.0 mm
B. 0.1 mm
C. 0.05 mm
D. 0.01 m

Answer

To solve the problem, we need to find the separation between the two coherent sources in a biprism experiment using the given parameters.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Fringe width, \( \beta = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Distance from the biprism to the screen, \( D = 1.0 \, \text{m} \)

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

3. **Rearrange the Formula to Solve for \( d \):**
We can rearrange the formula to find \( d \):
\[ d = \frac{D \lambda}{\beta} \]

4. **Substitute the Known Values:**
Now, substitute the values into the equation:
\[ d = \frac{1.0 \, \text{m} \times (5000 \times 10^{-10} \, \text{m})}{5 \times 10^{-3} \, \text{m}} \]

5. **Calculate the Separation \( d \):**
Performing the calculation:
\[ d = \frac{1.0 \times 5000 \times 10^{-10}}{5 \times 10^{-3}} = \frac{5000 \times 10^{-10}}{5 \times 10^{-3}} = \frac{5000}{5} \times 10^{-10 + 3} = 1000 \times 10^{-7} = 0.1 \times 10^{-3} \, \text{m} \]

6. **Convert to Millimeters:**
Convert the result to millimeters:
\[ d = 0.1 \, \text{mm} \]

7. **Final Result:**
The separation between the two coherent sources is \( 0.1 \, \text{mm} \).

### Conclusion:
The correct answer is \( 0.1 \, \text{mm} \).