The central fringe shifts to the position of fifth bright fringe, if a thin film of refractive index 1.5 is introduced in the path of light of wavelength `5000 Å`. The thickness of the glass plate is

Options

A. `1mum`
B. `5mum`
C. `3mum`
D. `4mum`

Answer

To solve the problem, we need to find the thickness of a glass plate that causes the central fringe in an interference pattern to shift to the position of the fifth bright fringe. Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The introduction of a thin film (glass plate) with a refractive index (μ) of 1.5 causes a shift in the interference pattern.
- The central fringe shifts to the position of the fifth bright fringe.

2. **Path Difference Calculation**:
- The path difference due to the introduction of the glass plate is given by:
\[ \Delta = (\mu - 1) \cdot t \]
- Where \( t \) is the thickness of the glass plate.

3. **Given Values**:
- Refractive index \( \mu = 1.5 \)
- Wavelength \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)

4. **Path Difference for the Fifth Bright Fringe**:
- The path difference for the nth bright fringe is given by:
\[ \Delta = n \cdot \lambda \]
- For the fifth bright fringe (n = 5):
\[ \Delta = 5 \cdot \lambda = 5 \cdot (5 \times 10^{-7}) = 25 \times 10^{-7} \, \text{m} \]

5. **Setting Up the Equation**:
- Since the central fringe shifts to the fifth bright fringe, we can equate the two expressions for path difference:
\[ (\mu - 1) \cdot t = 25 \times 10^{-7} \]
- Substituting \( \mu = 1.5 \):
\[ (1.5 - 1) \cdot t = 25 \times 10^{-7} \]
\[ 0.5 \cdot t = 25 \times 10^{-7} \]

6. **Solving for Thickness \( t \)**:
- Rearranging gives:
\[ t = \frac{25 \times 10^{-7}}{0.5} = 50 \times 10^{-7} \, \text{m} \]
- Converting to micrometers:
\[ t = 50 \times 10^{-7} \, \text{m} = 5 \, \mu m \]

### Final Answer:
The thickness of the glass plate is **5 micrometers**.

---