Young's double slit experiment is being performed with a light of wavelength 6000 Å. It is given that the path difference at the central bright fringe is zero, then what is the path difference between the light coming from different slits at the fourth bright fringe?

Options

A. `2.4 xx 10^(-6) m`
B. `1.2 xx 10^(-6) m`
C. `10^(-6) m `
D. `0.5 xx 10^(-6) m`

Explanation

To solve the problem, we need to determine the path difference between the light coming from the two slits at the fourth bright fringe in Young's double slit experiment.

### Step-by-Step Solution:

1. **Understanding the Concept of Bright Fringes**:
In Young's double slit experiment, bright fringes occur due to constructive interference of light waves. The condition for constructive interference (bright fringes) is given by:
\[ \Delta x = n \lambda \]
where \( \Delta x \) is the path difference, \( n \) is the order of the fringe (0, 1, 2, ...), and \( \lambda \) is the wavelength of the light used.

2. **Identifying the Wavelength**:
The problem states that the wavelength \( \lambda \) of the light used is 6000 Å (angstroms). We need to convert this into meters for standard SI units:
\[ 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

3. **Determining the Order of the Fringe**:
We are asked to find the path difference at the fourth bright fringe. For the fourth bright fringe, \( n = 4 \).

4. **Calculating the Path Difference**:
Using the formula for path difference:
\[ \Delta x = n \lambda \]
Substituting \( n = 4 \) and \( \lambda = 6 \times 10^{-7} \, \text{m} \):
\[ \Delta x = 4 \times (6 \times 10^{-7}) = 24 \times 10^{-7} \, \text{m} = 2.4 \times 10^{-6} \, \text{m} \]

5. **Final Answer**:
The path difference at the fourth bright fringe is:
\[ \Delta x = 2.4 \, \mu m \, \text{(micrometers)} \]

### Summary:
The path difference between the light coming from different slits at the fourth bright fringe is \( 2.4 \, \mu m \).