Two slits are separated by `2.00 xx 10^(-5)` m. They are illuminated by light of wavelength `5.60 xx 10^(-7)` m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?

Options

A. 0.421 m
B. 0.168 m
C. 0.224 m
D. 0.084 m

Answer

To find the separation between the central bright fringe and the third dark fringe in a double-slit interference pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Distance between the slits, \( d = 2.00 \times 10^{-5} \, \text{m} \)
- Wavelength of light, \( \lambda = 5.60 \times 10^{-7} \, \text{m} \)
- Distance from the slits to the screen, \( D = 6.00 \, \text{m} \)

2. **Understand the condition for dark fringes:**
The condition for dark fringes in a double-slit experiment is given by:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
where \( n \) is the order of the dark fringe. For the third dark fringe, \( n = 3 \).

3. **Calculate the path difference for the third dark fringe:**
Substituting \( n = 3 \) into the formula:
\[ \Delta x = (2 \cdot 3 - 1) \frac{\lambda}{2} = 5 \frac{\lambda}{2} \]

4. **Substitute the value of \( \lambda \):**
\[ \Delta x = 5 \cdot \frac{5.60 \times 10^{-7}}{2} = \frac{28.0 \times 10^{-7}}{2} = 14.0 \times 10^{-7} \, \text{m} \]

5. **Relate the path difference to the position of the dark fringe:**
The position of the dark fringe \( x_n \) on the screen can be found using:
\[ \Delta x = \frac{xd}{D} \]
Rearranging gives:
\[ x_n = \frac{\Delta x \cdot D}{d} \]

6. **Substituting the values:**
\[ x_3 = \frac{(14.0 \times 10^{-7}) \cdot 6.00}{2.00 \times 10^{-5}} \]

7. **Calculate \( x_3 \):**
\[ x_3 = \frac{8.4 \times 10^{-6}}{2.00 \times 10^{-5}} = 0.42 \, \text{m} \]

### Final Answer:
The separation between the central bright fringe and the third dark fringe is \( 0.42 \, \text{m} \).

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