The slits in Young's double slit experiment are illuminated by light of wavelength `6000 Å`. If the path difference at the central bright fringe is zero, what is the path difference for light from the slits at the fourth bright fringe?

Options

A. `2.4xx10^(-6)m`
B. `1.2xx10^(-6)m`
C. `10^(-6)m`
D. `0.5xx10^(-6)m`

Explanation

To solve the problem, we need to determine the path difference for the fourth bright fringe in Young's double slit experiment.

### Step-by-Step Solution:

1. **Understanding the Concept**:
In Young's double slit experiment, the path difference between the light waves from the two slits determines the position of the bright and dark fringes on the screen. The path difference for the nth bright fringe is given by the formula:
\[ \Delta x = n \lambda \]
where \( n \) is the fringe order (n = 0 for central maximum, n = 1 for the first bright fringe, n = 2 for the second bright fringe, and so on), and \( \lambda \) is the wavelength of the light used.

2. **Given Data**:
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- We need to find the path difference for the 4th bright fringe, which corresponds to \( n = 4 \).

3. **Calculating the Path Difference**:
Using the formula for path difference:
\[ \Delta x = n \lambda \]
Substituting \( n = 4 \) and \( \lambda = 6000 \times 10^{-10} \, \text{m} \):
\[ \Delta x = 4 \times (6000 \times 10^{-10} \, \text{m}) \]
\[ \Delta x = 24000 \times 10^{-10} \, \text{m} \]

4. **Converting to Scientific Notation**:
To express \( 24000 \times 10^{-10} \, \text{m} \) in scientific notation:
\[ 24000 = 2.4 \times 10^{4} \]
Thus,
\[ \Delta x = 2.4 \times 10^{4} \times 10^{-10} \, \text{m} = 2.4 \times 10^{-6} \, \text{m} \]

5. **Final Answer**:
The path difference for light from the slits at the fourth bright fringe is:
\[ \Delta x = 2.4 \times 10^{-6} \, \text{m} \]