On introducing a thin film in the path of one of the two interfering beam, the central fringe will shift by one fringe width. If `mu=1.5,` the thickness of the film is (wavelength of monochromatic light is `lambda`)

Options

A. `4lambda`
B. `3lambda`
C. `2lambda`
D. `lambda`

Answer

To solve the problem, we need to determine the thickness of a thin film that causes a shift of one fringe width in an interference pattern. We are given the refractive index (\( \mu \)) of the film and the wavelength (\( \lambda \)) of the monochromatic light.

### Step-by-Step Solution:

1. **Understanding the Shift in Fringes**:
When a thin film is introduced in the path of one of the interfering beams, it causes a change in the optical path length. This results in a shift in the interference pattern. The shift in the central fringe by one fringe width indicates that the path difference introduced by the film is equal to one wavelength (\( \lambda \)).

2. **Path Difference Due to the Thin Film**:
The path difference (\( \Delta x \)) introduced by a thin film of thickness \( t \) and refractive index \( \mu \) can be expressed as:
\[ \Delta x = (\mu - 1) t \]
This equation accounts for the additional optical path length due to the film.

3. **Setting Up the Equation**:
Since the central fringe shifts by one fringe width, we can set the path difference equal to one wavelength:
\[ (\mu - 1) t = n \lambda \]
where \( n \) is the number of fringes shifted. In this case, \( n = 1 \).

4. **Substituting Known Values**:
We know that \( \mu = 1.5 \) and \( n = 1 \). Substituting these values into the equation gives:
\[ (1.5 - 1) t = 1 \lambda \]
Simplifying, we have:
\[ 0.5 t = \lambda \]

5. **Solving for Thickness \( t \)**:
To find the thickness \( t \), we rearrange the equation:
\[ t = \frac{\lambda}{0.5} \]
This simplifies to:
\[ t = 2 \lambda \]

### Final Answer:
The thickness of the film is \( t = 2 \lambda \).