In Young.s double slit interference experiment the wavelength of light used is `6000 A^(0)`. If the path difference between waves reaching a point P on the screen is `1.5` microns, then at that point P

Options

A. second bright band occurs
B. second dark band occurs
C. third dark band occurs
D. third bright band occurs

Answer

To solve the problem, we need to determine the type of fringe (bright or dark) that occurs at point P in Young's double slit experiment given the path difference and the wavelength of light used.

### Step-by-Step Solution:

1. **Identify the given values**:
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Path difference, \( \Delta x = 1.5 \, \mu\text{m} = 1.5 \times 10^{-6} \, \text{m} \)

2. **Calculate the ratio of path difference to wavelength**:
\[ \frac{\Delta x}{\lambda} = \frac{1.5 \times 10^{-6} \, \text{m}}{6000 \times 10^{-10} \, \text{m}} \]

3. **Perform the calculation**:
- First, calculate \( 6000 \times 10^{-10} \):
\[ 6000 \times 10^{-10} = 6 \times 10^{-7} \, \text{m} \]
- Now substitute this back into the ratio:
\[ \frac{\Delta x}{\lambda} = \frac{1.5 \times 10^{-6}}{6 \times 10^{-7}} = \frac{1.5}{6} \times 10^{1} = 0.25 \times 10^{1} = 2.5 \]

4. **Interpret the result**:
- The ratio \( \frac{\Delta x}{\lambda} = 2.5 \) indicates that the path difference is \( 2.5 \) wavelengths.
- This can be expressed as:
\[ \Delta x = 2.5 \lambda \]

5. **Determine the type of fringe**:
- In Young's double slit experiment, a path difference of \( n\lambda \) (where \( n \) is an integer) corresponds to a bright fringe, while a path difference of \( (n + 0.5)\lambda \) corresponds to a dark fringe.
- Since \( 2.5 \) can be expressed as \( 2 + 0.5 \), it indicates that this is a dark fringe (specifically, the third dark fringe).

6. **Conclusion**:
- At point P, there is a third dark band (minimum) occurring.

### Final Answer:
At point P, there is a third dark band (minimum).

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