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

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 find the path difference for the fourth bright fringe in Young's double slit experiment given the wavelength of light.

### Step-by-step Solution:

1. **Identify the Wavelength**:
The wavelength of the light used in the experiment is given as 6000 Ångströms (Å). We need to convert this to meters for our calculations.
\[ \text{Wavelength} (\lambda) = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

2. **Understand Path Difference for Bright Fringes**:
The path difference (\( \Delta x \)) for the nth bright fringe in Young's double slit experiment is given by the formula:
\[ \Delta x = n \lambda \]
where \( n \) is the fringe number (n=1 for the first bright fringe, n=2 for the second, and so on).

3. **Determine the Value of n**:
For the fourth bright fringe, \( n = 4 \).

4. **Calculate the Path Difference**:
Substitute the values into the path difference formula:
\[ \Delta x_4 = 4 \lambda = 4 \times (6000 \times 10^{-10} \, \text{m}) \]
\[ \Delta x_4 = 4 \times 6000 \times 10^{-10} \, \text{m} = 24000 \times 10^{-10} \, \text{m} \]
\[ \Delta x_4 = 2.4 \times 10^{-6} \, \text{m} \]

5. **Final Result**:
The path difference for the light from the slits at the fourth bright fringe is:
\[ \Delta x_4 = 2.4 \times 10^{-6} \, \text{m} \]