In Young's double-slit experiment, the slits are `2mm` apart and are illuminated by photons of two wavelengths `lambda_1=12000Å` and `lambda_2=10000Å`. 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. (a) `8mm`
B. (b) `6mm`
C. (c) `4mm`
D. (d) `3mm`

Explanation

To solve the problem, we need to find the minimum distance from the common central bright fringe on the screen where a bright fringe from one interference pattern coincides with a bright fringe from the other. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Distance between the slits, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Wavelengths of the two light sources:
- \( \lambda_1 = 12000 \, \text{Å} = 12000 \times 10^{-10} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \)
- \( \lambda_2 = 10000 \, \text{Å} = 10000 \times 10^{-10} \, \text{m} = 1.0 \times 10^{-6} \, \text{m} \)
- Distance from the slits to the screen, \( D = 2 \, \text{m} \)

2. **Fringe Position Formula:**
The position of the bright fringes in Young's double-slit experiment is given by:
\[ X_n = \frac{n \lambda D}{d} \]
where \( n \) is the order of the fringe.

3. **Set Up the Coincidence Condition:**
For the bright fringes from both wavelengths to coincide:
\[ X_{n_1} = X_{n_2} \]
This gives us:
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
Since \( D \) and \( d \) are the same for both equations, they cancel out:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

4. **Express the Ratio of Orders:**
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]
Substituting the values:
\[ \frac{n_1}{n_2} = \frac{1.0 \times 10^{-6}}{1.2 \times 10^{-6}} = \frac{10}{12} = \frac{5}{6} \]
This implies \( n_1 = 5k \) and \( n_2 = 6k \) for some integer \( k \).

5. **Find Minimum Values of \( n_1 \) and \( n_2 \):**
The smallest integers satisfying \( n_1 = 5 \) and \( n_2 = 6 \) are \( n_1 = 5 \) and \( n_2 = 6 \).

6. **Calculate the Position \( X_n \):**
Using \( n_1 = 5 \):
\[ X_{n_1} = \frac{5 \cdot \lambda_1 \cdot D}{d} \]
Substituting the values:
\[ X_{n_1} = \frac{5 \cdot (1.2 \times 10^{-6}) \cdot 2}{2 \times 10^{-3}} = \frac{5 \cdot 1.2 \times 10^{-6} \cdot 2}{2 \times 10^{-3}} = \frac{5 \cdot 1.2 \times 10^{-6}}{10^{-3}} = 5 \cdot 1.2 \times 10^{-3} = 6 \times 10^{-3} \, \text{m} = 6 \, \text{mm} \]

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