In Young's double-slit experiment, 30 fringes are obtained in the field of view of the observing telescope, when the wavelength of light used is `4000 Å`. If we use monochromatic light of wavelength `6000 Å`, the number of fringes obtained in the same field of view is

Options

A. 30
B. 45
C. 20
D. none of these

Explanation

To solve the problem of finding the number of fringes obtained in Young's double-slit experiment when the wavelength of light changes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the number of fringes and wavelength**:
In Young's double-slit experiment, the number of fringes (n) is inversely proportional to the wavelength (λ) of the light used. This can be expressed mathematically as:
\[ n \propto \frac{1}{\lambda} \]
This means that if the wavelength increases, the number of fringes decreases.

2. **Set up the equation using the given information**:
We can set up the relationship between the number of fringes and the wavelengths for two different cases:
\[ n_1 \cdot \lambda_1 = n_2 \cdot \lambda_2 \]
where:
- \( n_1 \) = number of fringes with wavelength \( \lambda_1 \)
- \( n_2 \) = number of fringes with wavelength \( \lambda_2 \)

3. **Substitute the known values**:
From the problem, we know:
- \( n_1 = 30 \) (for \( \lambda_1 = 4000 \, \text{Å} \))
- \( \lambda_1 = 4000 \, \text{Å} \)
- \( \lambda_2 = 6000 \, \text{Å} \)

Plugging these values into the equation gives:
\[ 30 \cdot 4000 = n_2 \cdot 6000 \]

4. **Solve for \( n_2 \)**:
Rearranging the equation to solve for \( n_2 \):
\[ n_2 = \frac{30 \cdot 4000}{6000} \]

5. **Calculate \( n_2 \)**:
Performing the calculation:
\[ n_2 = \frac{120000}{6000} = 20 \]

6. **Conclusion**:
Therefore, the number of fringes obtained when using monochromatic light of wavelength \( 6000 \, \text{Å} \) is \( n_2 = 20 \).

### Final Answer:
The number of fringes obtained in the same field of view with a wavelength of \( 6000 \, \text{Å} \) is **20**.

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