In Young.s double slit experiment with monochromatic source of light of wavelength `6000 A^(2)`, if the path difference at a point on the screen is `6xx 10^(-6)m`, the number of the bright band formed at that point is

Options

A. `2`
B. `4`
C. `6`
D. `10`

Answer

To solve the problem of finding the number of bright bands formed at a point on the screen in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

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

2. **Use the formula for path difference**:
The path difference for bright fringes in Young's double slit experiment is given by:
\[ \Delta x = n \lambda \]
where \( n \) is the order of the bright fringe.

3. **Rearrange the formula to find \( n \)**:
We can rearrange the formula to solve for \( n \):
\[ n = \frac{\Delta x}{\lambda} \]

4. **Substitute the known values**:
Substitute the values of \( \Delta x \) and \( \lambda \) into the equation:
\[ n = \frac{6 \times 10^{-6} \, \text{m}}{6000 \times 10^{-10} \, \text{m}} \]

5. **Calculate \( n \)**:
- First, simplify the denominator:
\[ 6000 \times 10^{-10} = 6 \times 10^{-7} \, \text{m} \]
- Now substitute this back into the equation:
\[ n = \frac{6 \times 10^{-6}}{6 \times 10^{-7}} = \frac{6}{6} \times \frac{10^{-6}}{10^{-7}} = 1 \times 10^{1} = 10 \]

6. **Conclusion**:
The number of bright bands formed at that point is \( n = 10 \).

### Final Answer:
The number of the bright band formed at that point is **10**.

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