In a given region of 10 fringes are observed in the reflected beam from a thin film. If the wavelength of th incident light is changed from `4200 Å " to " 6000 Å` , then the number of fringes observed in the same region will be

Options

A. 4
B. 7
C. 11
D. 3

Answer

To solve the problem, we need to determine how the number of fringes observed in the reflected beam from a thin film changes when the wavelength of the incident light is altered.

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
We start with 10 fringes observed with a wavelength of \( \lambda_1 = 4200 \, \text{Å} \).

2. **Use the Relationship Between Fringes and Wavelength**:
The number of fringes \( n \) is related to the wavelength \( \lambda \) and the thickness \( T \) of the film by the equation:
\[ n = \frac{2 \mu T}{\lambda} \]
where \( \mu \) is the refractive index of the film.

3. **Establish the Constant**:
Since the thickness \( T \) and refractive index \( \mu \) remain constant, we can set up the relationship for two different wavelengths:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
where \( n_1 = 10 \), \( \lambda_1 = 4200 \, \text{Å} \), and \( \lambda_2 = 6000 \, \text{Å} \).

4. **Substitute the Known Values**:
Plugging in the values we have:
\[ 10 \times 4200 = n_2 \times 6000 \]

5. **Solve for \( n_2 \)**:
Rearranging the equation to find \( n_2 \):
\[ n_2 = \frac{10 \times 4200}{6000} \]

6. **Calculate \( n_2 \)**:
Performing the calculation:
\[ n_2 = \frac{42000}{6000} = 7 \]

7. **Conclusion**:
Therefore, when the wavelength is changed from \( 4200 \, \text{Å} \) to \( 6000 \, \text{Å} \), the number of fringes observed in the same region will be \( 7 \).

### Final Answer:
The number of fringes observed in the same region will be **7**.