In the Young's double slit experiment, maximum of bright bands observed (inclusive of the central bright band) is found to be 11. If `lambda` is the wavelength of the monochromatic light used, the distance between the slits is

Options

A. `5 lambda`
B. `6 lambda`
C. `10 lambda`
D. `11 lambda`

Explanation

To solve the problem, we need to determine the distance between the slits (denoted as \( d \)) in the Young's double slit experiment given that the maximum number of bright bands observed (including the central bright band) is 11.

### Step-by-Step Solution:

1. **Understanding the Number of Bright Bands**:
- The total number of bright bands observed is 11, which includes the central bright band.
- This means there are 5 bright bands on one side of the central band and 5 on the other side.

2. **Identifying the Value of \( n \)**:
- In the context of the interference pattern, \( n \) represents the order of the bright fringe. Since we have 5 bright fringes on either side of the central band, the maximum order \( n \) for the bright fringes is 5.

3. **Using the Formula for Path Difference**:
- The path difference \( \Delta \) for bright fringes is given by:
\[ \Delta = n \lambda \]
- Here, \( \lambda \) is the wavelength of the light used.

4. **Relating Path Difference to the Geometry of the Experiment**:
- The path difference can also be expressed in terms of the angle \( \theta \) and the distance between the slits \( d \):
\[ \Delta = d \sin \theta \]
- For maximum bright fringe visibility, we consider the angle \( \theta \) to be approximately 90 degrees for the last fringe, where \( \sin 90^\circ = 1 \).

5. **Setting Up the Equation**:
- Combining the two expressions for path difference, we have:
\[ n \lambda = d \sin \theta \]
- Substituting \( \sin \theta = 1 \) when \( \theta \) is 90 degrees:
\[ n \lambda = d \]

6. **Calculating the Distance Between the Slits**:
- Since we found \( n = 5 \):
\[ d = 5 \lambda \]

### Final Answer:
The distance between the slits \( d \) is \( 5 \lambda \).

---