In Young,s double slit experimentm, 60 fringes are observed in the central view zone with light of wavelength `4000Å`, The number of fringes that will be observed in the same view zone with the light of wavelength `6000Å`, is

Options

A. 40
B. 90
C. 60
D. None of these

Explanation

To solve the problem of finding the number of fringes observed in the central view zone of Young's double slit experiment with a different wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Wavelength 1, \( \lambda_1 = 4000 \, \text{Å} \)
- Number of fringes with \( \lambda_1 \), \( n_1 = 60 \)
- Wavelength 2, \( \lambda_2 = 6000 \, \text{Å} \)
- Number of fringes with \( \lambda_2 \), \( n_2 = ? \)

2. **Fringe Width Formula**:
- The fringe width \( w \) in Young's double slit experiment is given by:
\[ w = \frac{\lambda D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

3. **Total Width of the Screen**:
- The total width of the screen \( A \) can be expressed in terms of the number of fringes and fringe width:
\[ A = n \cdot w \]
- For the first case (with \( \lambda_1 \)):
\[ A = n_1 \cdot w_1 = n_1 \cdot \frac{\lambda_1 D}{d} \]

4. **Setting Up the Equation for Both Cases**:
- For the second case (with \( \lambda_2 \)):
\[ A = n_2 \cdot w_2 = n_2 \cdot \frac{\lambda_2 D}{d} \]
- Since the width of the screen \( A \) is the same in both cases, we can set the two equations equal:
\[ n_1 \cdot \frac{\lambda_1 D}{d} = n_2 \cdot \frac{\lambda_2 D}{d} \]

5. **Simplifying the Equation**:
- The terms \( D \) and \( d \) cancel out, leading to:
\[ n_1 \cdot \lambda_1 = n_2 \cdot \lambda_2 \]

6. **Rearranging to Find \( n_2 \)**:
- Rearranging gives us:
\[ n_2 = \frac{n_1 \cdot \lambda_1}{\lambda_2} \]

7. **Substituting the Values**:
- Substitute \( n_1 = 60 \), \( \lambda_1 = 4000 \, \text{Å} \), and \( \lambda_2 = 6000 \, \text{Å} \):
\[ n_2 = \frac{60 \cdot 4000}{6000} \]

8. **Calculating \( n_2 \)**:
- Simplifying the expression:
\[ n_2 = \frac{240000}{6000} = 40 \]

### Final Answer:
The number of fringes observed with the wavelength \( 6000 \, \text{Å} \) is \( n_2 = 40 \).