In Young's double slit experiment, we get `60` fringes in the field of view of monochromatic light of wavelength `4000Å`. If we use monochromatic light of wavelength `6000Å`, then the number of fringes obtained in the same field of view is

Options

A. (a) `60`
B. (b) `90`
C. (c) `40`
D. (d) `1.5`

Explanation

To solve the problem, we need to find the number of fringes obtained when the wavelength of the monochromatic light is changed from \(4000 \, \text{Å}\) to \(6000 \, \text{Å}\).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
In Young's double slit experiment, the number of fringes \(N\) in a given field of view is given by the formula:
\[ N = \frac{L}{\beta} \]
where \(L\) is the length of the region and \(\beta\) is the fringe width.

2. **Fringe Width Calculation**:
The fringe width \(\beta\) is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits.

3. **Using the Given Values**:
For the first case with wavelength \(\lambda_1 = 4000 \, \text{Å}\) and \(N_1 = 60\):
\[ \beta_1 = \frac{\lambda_1 D}{d} = \frac{4000 \, \text{Å} \cdot D}{d} \]
Thus, the number of fringes can be expressed as:
\[ N_1 = \frac{L}{\beta_1} = \frac{L \cdot d}{4000 \, \text{Å} \cdot D} \]

4. **Finding the Length of the Region**:
Rearranging gives us:
\[ L = N_1 \cdot \beta_1 = 60 \cdot \frac{4000 \, \text{Å} \cdot D}{d} \]

5. **Calculating for the New Wavelength**:
Now, for the second case with wavelength \(\lambda_2 = 6000 \, \text{Å}\):
\[ \beta_2 = \frac{\lambda_2 D}{d} = \frac{6000 \, \text{Å} \cdot D}{d} \]
The number of fringes \(N_2\) can be expressed as:
\[ N_2 = \frac{L}{\beta_2} = \frac{L \cdot d}{6000 \, \text{Å} \cdot D} \]

6. **Substituting for Length \(L\)**:
Substitute \(L\) from step 4 into the equation for \(N_2\):
\[ N_2 = \frac{60 \cdot \frac{4000 \, \text{Å} \cdot D}{d} \cdot d}{6000 \, \text{Å} \cdot D} \]
Simplifying this gives:
\[ N_2 = \frac{60 \cdot 4000 \, \text{Å}}{6000 \, \text{Å}} \]

7. **Final Calculation**:
\[ N_2 = 60 \cdot \frac{4000}{6000} = 60 \cdot \frac{2}{3} = 40 \]

### Conclusion:
The number of fringes obtained in the same field of view with a wavelength of \(6000 \, \text{Å}\) is \(40\).