In Young.s double slit experiment, the slits are 2mm apart and are illuminated by photons of two wavelengths `lambda_(1)= 12000 A^(0)" and "lambda_(2)= 10,000 A^(0)`. At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options

A. `3mm`
B. `8mm`
C. `6mm`
D. `4mm`

Answer

To solve the problem, we need to determine the minimum distance from the central bright fringe on the screen where a bright fringe from one wavelength coincides with a bright fringe from the other wavelength in Young's double slit experiment.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelengths:
- \( \lambda_1 = 12000 \, \text{Å} = 12000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 10000 \, \text{Å} = 10000 \times 10^{-10} \, \text{m} \)
- Distance between slits:
- \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen:
- \( D = 2 \, \text{m} \)

2. **Calculate Fringe Width:**
The fringe width \( \beta \) for each wavelength is given by the formula:
\[ \beta = \frac{d \lambda}{D} \]
- For \( \lambda_1 \):
\[ \beta_1 = \frac{d \lambda_1}{D} = \frac{(2 \times 10^{-3}) \times (12000 \times 10^{-10})}{2} = \frac{2 \times 12000 \times 10^{-13}}{2} = 12000 \times 10^{-13} \, \text{m} \]
- For \( \lambda_2 \):
\[ \beta_2 = \frac{d \lambda_2}{D} = \frac{(2 \times 10^{-3}) \times (10000 \times 10^{-10})}{2} = \frac{2 \times 10000 \times 10^{-13}}{2} = 10000 \times 10^{-13} \, \text{m} \]

3. **Determine the Condition for Coincidence:**
The bright fringes coincide when:
\[ n_1 \beta_1 = n_2 \beta_2 \]
Substituting the values of \( \beta_1 \) and \( \beta_2 \):
\[ n_1 (12000 \times 10^{-13}) = n_2 (10000 \times 10^{-13}) \]
Simplifying gives:
\[ \frac{n_1}{n_2} = \frac{10000}{12000} = \frac{5}{6} \]

4. **Choose Values of \( n_1 \) and \( n_2 \):**
Let \( n_1 = 5k \) and \( n_2 = 6k \) for some integer \( k \). The smallest values occur when \( k = 1 \):
- \( n_1 = 5 \)
- \( n_2 = 6 \)

5. **Calculate the Position of Coinciding Bright Fringes:**
The position \( x \) of the bright fringe on the screen is given by:
\[ x = n \beta D \]
- For \( n_1 = 5 \) (using \( \beta_1 \)):
\[ x_1 = 5 \cdot (12000 \times 10^{-13}) \cdot 2 = 5 \cdot 12000 \cdot 10^{-13} \cdot 2 = 120000 \times 10^{-13} \, \text{m} = 0.012 \, \text{m} = 12 \, \text{mm} \]
- For \( n_2 = 6 \) (using \( \beta_2 \)):
\[ x_2 = 6 \cdot (10000 \times 10^{-13}) \cdot 2 = 6 \cdot 10000 \cdot 10^{-13} \cdot 2 = 120000 \times 10^{-13} \, \text{m} = 0.012 \, \text{m} = 12 \, \text{mm} \]

6. **Minimum Distance from Central Bright Fringe:**
The minimum distance from the central bright fringe where the bright fringes coincide is:
\[ x = 6 \, \text{mm} \]

### Final Answer:
The minimum distance from the common central bright fringe where a bright fringe from one interference pattern coincides with a bright fringe from the other is **6 mm**.