In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength `6000Å`, coming from the coherent sources `S_1` and `S_2`. At certain point P on the screen third dark fringe is formed. Then the path difference `S_1P-S_2P` in microns is

Options

A. `0.75`
B. `1.5`
C. `3.0`
D. `4.5`

Explanation

To solve the problem, we need to determine the path difference \( S_1P - S_2P \) at the point where the third dark fringe is formed in Young's double slit experiment.

### Step-by-Step Solution:

1. **Identify the Wavelength**:
The wavelength of the light used is given as \( \lambda = 6000 \, \text{Å} \).
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6000 \times 10^{-4} \, \mu m = 0.6 \, \mu m \]

2. **Determine the Order of the Dark Fringe**:
We are looking for the third dark fringe, which corresponds to \( n = 3 \).

3. **Use the Formula for Path Difference**:
The path difference for dark fringes in Young's double slit experiment is given by:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
Substituting \( n = 3 \):
\[ \Delta x = (2 \times 3 - 1) \frac{\lambda}{2} = (6 - 1) \frac{\lambda}{2} = 5 \frac{\lambda}{2} \]

4. **Substitute the Wavelength**:
Now, substitute \( \lambda = 6000 \, \text{Å} \):
\[ \Delta x = 5 \frac{6000 \, \text{Å}}{2} = 5 \times 3000 \, \text{Å} = 15000 \, \text{Å} \]

5. **Convert to Microns**:
To convert \( 15000 \, \text{Å} \) to microns:
\[ 15000 \, \text{Å} = 15000 \times 10^{-10} \, \text{m} = 15000 \times 10^{-4} \, \mu m = 1.5 \, \mu m \]

6. **Final Result**:
Therefore, the path difference \( S_1P - S_2P \) is:
\[ \Delta x = 1.5 \, \mu m \]

### Conclusion:
The path difference \( S_1P - S_2P \) at the point where the third dark fringe is formed is \( 1.5 \, \mu m \).

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