A screen is at a distance of 2m that are narrow slit illuminated with light of `6000 A^(@)`. The first maximum lies at 0.005mm on either side of the central maximum, then the distance between the slits will be

Options

A. 0.024 mm
B. 0.24 mm
C. 2.4 mm
D. 24 mm

Answer

To solve the problem, we need to find the distance between the slits (denoted as \( d \)) given the distance to the screen, the wavelength of the light, and the position of the first maximum.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Distance to the screen (\( D \)) = 2 m
- Wavelength of light (\( \lambda \)) = 6000 Å = \( 6000 \times 10^{-10} \) m
- Position of the first maximum from the central maximum (\( y_1 \)) = 0.005 mm = \( 0.005 \times 10^{-3} \) m

2. **Convert Units**:
- Convert the position of the first maximum to meters:
\[ y_1 = 0.005 \text{ mm} = 0.005 \times 10^{-3} \text{ m} = 5 \times 10^{-6} \text{ m} \]

3. **Use the Formula for Fringe Width**:
The formula for the position of the first maximum in a double-slit experiment is given by:
\[ y_n = \frac{n \lambda D}{d} \]
where \( n \) is the order of the maximum (for the first maximum, \( n = 1 \)).

4. **Rearranging the Formula**:
To find the distance between the slits (\( d \)), we rearrange the formula:
\[ d = \frac{n \lambda D}{y_n} \]

5. **Substituting the Known Values**:
Substitute \( n = 1 \), \( \lambda = 6000 \times 10^{-10} \) m, \( D = 2 \) m, and \( y_n = 5 \times 10^{-6} \) m into the formula:
\[ d = \frac{1 \times (6000 \times 10^{-10}) \times 2}{5 \times 10^{-6}} \]

6. **Calculating the Distance Between the Slits**:
\[ d = \frac{12000 \times 10^{-10}}{5 \times 10^{-6}} = \frac{12000}{5} \times 10^{-4} = 2400 \times 10^{-4} = 0.24 \text{ m} \]

### Final Answer:
The distance between the slits is \( d = 0.24 \text{ m} \).