Separation between the slits in Young's double-slit experiment is 0.2 mm and separation between plane of the slits and screen is 2m. Wavelength of light used in the experiment is `5000 Å`. If first maximum is obtained at a distance x from the centre then what is x in mm?
`{:(0,1,2,3,4,5,6,7,8,9):}`

Explanation

To solve the problem, we will use the formula for the position of the bright fringes in Young's double-slit experiment. The formula for the position of the nth maximum is given by:

\[ y_n = \frac{n \lambda D}{d} \]

where:
- \(y_n\) is the distance from the central maximum to the nth maximum,
- \(n\) is the order of the maximum (for the first maximum, \(n = 1\)),
- \(\lambda\) is the wavelength of the light used,
- \(D\) is the distance from the slits to the screen,
- \(d\) is the separation between the slits.

### Step 1: Convert the given values into appropriate units
1. **Separation between the slits (\(d\))**:
- Given \(d = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} = 2 \times 10^{-4} \, \text{m}\).

2. **Distance from slits to screen (\(D\))**:
- Given \(D = 2 \, \text{m}\).

3. **Wavelength of light (\(\lambda\))**:
- Given \(\lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m}\).

### Step 2: Substitute the values into the formula
Using the formula for the first maximum (\(n = 1\)):
\[ y_1 = \frac{1 \cdot \lambda \cdot D}{d} \]
Substituting the values:
\[ y_1 = \frac{1 \cdot (5 \times 10^{-7} \, \text{m}) \cdot (2 \, \text{m})}{2 \times 10^{-4} \, \text{m}} \]

### Step 3: Calculate \(y_1\)
\[ y_1 = \frac{5 \times 10^{-7} \cdot 2}{2 \times 10^{-4}} = \frac{10 \times 10^{-7}}{2 \times 10^{-4}} = \frac{10}{2} \times 10^{-3} = 5 \times 10^{-3} \, \text{m} \]

### Step 4: Convert \(y_1\) from meters to millimeters
To convert from meters to millimeters:
\[ y_1 = 5 \times 10^{-3} \, \text{m} = 5 \, \text{mm} \]

### Final Answer
The distance \(x\) from the center to the first maximum is:
\[ \boxed{5 \, \text{mm}} \]