In young's double-slit experiment, the slit are 2 mm apart and are illuminated with a mixture ot two wavelengths `lambda_(0) = 750 nm` and `lambda = 900 nm`, The minimum distance from the common central bright fringe on a screen 2 m from the slits, where a bright fringe from one interference pattern coincides with a bright fringe from the other, is

Options

A. 1.5 mm
B. 3 mm
C. 4.5 mm
D. 6 mm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We need to find the minimum distance from the central bright fringe on the screen where a bright fringe from the interference pattern of two different wavelengths coincides.

### Step 2: Identify given values
- Distance between slits (d) = 2 mm = \(2 \times 10^{-3}\) m
- Distance from slits to screen (D) = 2 m
- Wavelength 1 (\(\lambda_1\)) = 750 nm = \(750 \times 10^{-9}\) m
- Wavelength 2 (\(\lambda_2\)) = 900 nm = \(900 \times 10^{-9}\) m

### Step 3: Set up the relationship for fringe positions
For Young's double-slit experiment, the position of the bright fringes on the screen can be given by the formula:
\[ y = \frac{n \lambda D}{d} \]
where \(y\) is the position of the nth bright fringe, \(n\) is the order of the fringe, \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the distance between the slits.

Let \(n_1\) be the order of the fringe for wavelength \(\lambda_1\) and \(n_2\) be the order of the fringe for wavelength \(\lambda_2\). The positions of the bright fringes can be expressed as:
\[ y_1 = \frac{n_1 \lambda_1 D}{d} \]
\[ y_2 = \frac{n_2 \lambda_2 D}{d} \]

### Step 4: Set the fringe positions equal to find the condition for coincidence
To find the minimum distance where the fringes coincide:
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
This simplifies to:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
From this, we can express the ratio of \(n_1\) and \(n_2\):
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]

### Step 5: Substitute the wavelengths
Substituting the values of the wavelengths:
\[ \frac{n_1}{n_2} = \frac{900 \times 10^{-9}}{750 \times 10^{-9}} = \frac{900}{750} = \frac{6}{5} \]
This means:
\[ n_1 = 6k \quad \text{and} \quad n_2 = 5k \]
for some integer \(k\).

### Step 6: Find the position of the coinciding fringe
Substituting \(n_1\) into the equation for \(y_1\):
\[ y_1 = \frac{6k \cdot 750 \times 10^{-9} \cdot 2}{2 \times 10^{-3}} = \frac{6k \cdot 750 \times 10^{-9} \cdot 2}{2 \times 10^{-3}} = 6k \cdot 750 \times 10^{-6} \]
Calculating for \(k=1\):
\[ y_1 = 6 \cdot 750 \times 10^{-6} = 4500 \times 10^{-6} = 4.5 \text{ mm} \]

### Conclusion
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 is **4.5 mm**.