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 `lamda`)

Options

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

Answer

To solve the problem, we need to determine the thickness of the thin film that causes the central fringe to shift by one fringe width when the refractive index (µ) is given as 1.5 and the wavelength of the monochromatic light is λ.

### Step-by-Step Solution:

1. **Understanding the Shift in Interference Pattern**:
When a thin film is introduced in the path of one of the interfering beams, it alters the optical path length. The shift in the interference pattern can be quantified by the formula:
\[ \text{Shift} = \frac{(µ - 1) \cdot t \cdot D}{d} \]
where:
- \( µ \) = refractive index of the film,
- \( t \) = thickness of the film,
- \( D \) = distance between the slits and the screen,
- \( d \) = distance between the slits.

2. **Setting Up the Condition for One Fringe Width Shift**:
Since the problem states that the central fringe shifts by one fringe width, we can set the shift equal to the fringe width \( \beta \):
\[ \beta = \frac{\lambda D}{d} \]
Therefore, we can equate the two expressions:
\[ \frac{(µ - 1) \cdot t \cdot D}{d} = \frac{\lambda D}{d} \]

3. **Cancelling Common Terms**:
We can cancel \( D \) and \( d \) from both sides (assuming they are not zero):
\[ (µ - 1) \cdot t = \lambda \]

4. **Substituting the Given Values**:
Now, substituting \( µ = 1.5 \):
\[ (1.5 - 1) \cdot t = \lambda \]
Simplifying this gives:
\[ 0.5 \cdot t = \lambda \]

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

Thus, the thickness of the film is:
\[ \boxed{2\lambda} \]