In Young's double slit experiment, we get 10 fringes in the field of view of monochromatic light of wavelength 4000Å. If we use monochromatic light of wavelength 5000Å, then the number of fringes obtained in the same field of view is

Options

A. 8
B. 10
C. 40
D. 50

Explanation

To solve the problem, we need to determine the number of fringes obtained in Young's double slit experiment when the wavelength of the light changes from 4000 Å to 5000 Å.

### Step-by-Step Solution:

1. **Understanding the Relationship**: In Young's double slit experiment, the number of fringes (N) observed in a given field of view is related to the wavelength (λ) of the light used. The formula that relates these quantities is:
\[ N \propto \frac{1}{\lambda} \]
This means that the number of fringes is inversely proportional to the wavelength.

2. **Setting Up the Initial Condition**: We know that with a wavelength of 4000 Å, the number of fringes (N1) is 10:
\[ N_1 = 10 \quad \text{and} \quad \lambda_1 = 4000 \, \text{Å} \]

3. **Finding the New Wavelength**: We need to find the number of fringes (N2) when the wavelength is changed to 5000 Å:
\[ \lambda_2 = 5000 \, \text{Å} \]

4. **Using the Proportionality**: Since the number of fringes is inversely proportional to the wavelength, we can set up the following relationship:
\[ \frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1} \]

5. **Substituting Known Values**: Plugging in the known values:
\[ \frac{10}{N_2} = \frac{5000}{4000} \]

6. **Cross-Multiplying**: To solve for N2, we cross-multiply:
\[ 10 \cdot 4000 = N_2 \cdot 5000 \]

7. **Calculating N2**: Simplifying the equation gives:
\[ N_2 = \frac{10 \cdot 4000}{5000} \]
\[ N_2 = \frac{40000}{5000} = 8 \]

8. **Conclusion**: Therefore, the number of fringes obtained with a wavelength of 5000 Å in the same field of view is:
\[ N_2 = 8 \]

### Final Answer:
The number of fringes obtained with a wavelength of 5000 Å is **8**.

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