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 step by step, we will use the relationship between the number of fringes, the wavelength of light, and the length of the region being observed.

### Step 1: Understand the relationship between the number of fringes and wavelength
We know that the number of fringes (n) observed in a certain region is given by the formula:
\[ n \times w = l \]
where \( w \) is the fringe width and \( l \) is the length of the region.

### Step 2: Express fringe width in terms of wavelength
The fringe width \( w \) can be expressed as:
\[ w = \frac{\lambda D}{t} \]
where \( \lambda \) is the wavelength of light, \( D \) is the distance from the film to the screen, and \( t \) is the thickness of the film.

### Step 3: Set up the equations for two different wavelengths
For the first case (with wavelength \( \lambda_1 = 4200 \, \text{Å} \)):
\[ n_1 \times w_1 = l \]
For the second case (with wavelength \( \lambda_2 = 6000 \, \text{Å} \)):
\[ n_2 \times w_2 = l \]

Since \( l \) is constant, we can equate the two expressions:
\[ n_1 \times w_1 = n_2 \times w_2 \]

### Step 4: Substitute the expression for fringe width
Substituting the expression for fringe width into the equation:
\[ n_1 \times \left( \frac{\lambda_1 D}{t} \right) = n_2 \times \left( \frac{\lambda_2 D}{t} \right) \]

### Step 5: Cancel common terms
Since \( D \) and \( t \) are constants, they cancel out:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

### Step 6: Substitute known values
Given that \( n_1 = 10 \), \( \lambda_1 = 4200 \, \text{Å} \), and \( \lambda_2 = 6000 \, \text{Å} \):
\[ 10 \times 4200 = n_2 \times 6000 \]

### Step 7: Solve for \( n_2 \)
Now, we can solve for \( n_2 \):
\[ n_2 = \frac{10 \times 4200}{6000} \]
\[ n_2 = \frac{42000}{6000} \]
\[ n_2 = 7 \]

### Conclusion
The number of fringes observed in the same region when the wavelength changes to \( 6000 \, \text{Å} \) is **7**.

---