A central fringe of interference pattern produced by light of wavelength `6000 Å` is shifted to the position of 5th bright fringe by introducing thin film of `mu = 1.5`. Calculate thickness of the film.

Explanation

To solve the problem, we need to calculate the thickness of a thin film that causes a central fringe of an interference pattern to shift to the position of the 5th bright fringe. The given data includes the wavelength of light and the refractive index of the film.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Refractive index of the film, \( \mu = 1.5 \)
- Shift to the position of the 5th bright fringe, \( N = 5 \)

2. **Understand the Concept of Fringe Shift:**
- When a thin film is introduced, it causes a phase change in the light passing through it. The path difference introduced by the film can be expressed as:
\[ \text{Path difference} = (\mu - 1) \cdot T \]
where \( T \) is the thickness of the film.

3. **Relate Path Difference to Fringe Shift:**
- The path difference corresponding to a shift of \( N \) bright fringes is given by:
\[ \text{Path difference} = N \cdot \lambda \]

4. **Set Up the Equation:**
- Equating the two expressions for path difference, we have:
\[ (\mu - 1) \cdot T = N \cdot \lambda \]
- Substituting the known values:
\[ (1.5 - 1) \cdot T = 5 \cdot (6 \times 10^{-7}) \]
- Simplifying the left side:
\[ 0.5 \cdot T = 5 \cdot 6 \times 10^{-7} \]

5. **Calculate the Right Side:**
- Calculate \( 5 \cdot 6 \):
\[ 5 \cdot 6 = 30 \]
- Thus, we have:
\[ 0.5 \cdot T = 30 \times 10^{-7} \]

6. **Solve for Thickness \( T \):**
- To find \( T \), divide both sides by \( 0.5 \):
\[ T = \frac{30 \times 10^{-7}}{0.5} = 60 \times 10^{-7} \, \text{m} \]
- Converting to a more standard form:
\[ T = 6 \times 10^{-6} \, \text{m} \]

### Final Answer:
The thickness of the film is \( T = 6 \times 10^{-6} \, \text{m} \) or \( 6 \, \mu m \).

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