In Young's double slit experiment, if the distance between two slits is equal to the wavelength of used light. Then the maximum number of bright firnges obtained on the screen will be

Options

A. Infinite
B. 3
C. 7
D. 5

Answer

To solve the problem of finding the maximum number of bright fringes in Young's double slit experiment when the distance between the two slits is equal to the wavelength of the light used, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In Young's double slit experiment, we have two slits separated by a distance \( d \).
- The wavelength of the light used is denoted as \( \lambda \).

2. **Using the Path Difference Condition**:
- The path difference \( \Delta x \) for the light coming from the two slits to a point on the screen is given by:
\[ \Delta x = d \sin \theta \]
- For constructive interference (bright fringes), this path difference must be an integer multiple of the wavelength:
\[ \Delta x = n \lambda \quad (n = 0, 1, 2, \ldots) \]

3. **Setting Up the Equation**:
- From the above two equations, we can equate:
\[ d \sin \theta = n \lambda \]
- Rearranging gives:
\[ \sin \theta = \frac{n \lambda}{d} \]

4. **Substituting the Given Condition**:
- Given that \( d = \lambda \), we substitute this into our equation:
\[ \sin \theta = \frac{n \lambda}{\lambda} = n \]

5. **Finding the Maximum Value of \( n \)**:
- The sine function has a maximum value of 1. Therefore, for the maximum number of bright fringes:
\[ n \leq 1 \]
- This means \( n \) can take values 0 and 1.

6. **Counting the Bright Fringes**:
- The values of \( n \) that correspond to bright fringes are:
- \( n = 0 \) (central maximum)
- \( n = 1 \) (first order maximum)
- Thus, there are a total of 2 bright fringes.

### Conclusion:
The maximum number of bright fringes obtained on the screen is **2**.