A glass sheet `12xx10^(-3)` mm thick `(mu_(g)=1.5)` is placed in the path of one of the interfering beams in a Yongs's double slit interference arrangement using monochromatic light of wavelenght 6000 A. If the central bright fringe shifts a distance equal to width of 10 bands. What is the thickness of the sheet of diamond of refractive index 2.5 that has to be introduced in the path of second beam to bring th ecentral bright fringe to original position?

Options

A. `20//9 mum`
B. `7.2 mum`
C. `3 mum`
D. `4 mum`

Explanation

To solve the problem, we need to determine the thickness of a diamond sheet that will compensate for the shift caused by the glass sheet in a Young's double-slit interference arrangement.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Thickness of the glass sheet, \( T_g = 12 \times 10^{-3} \) mm = \( 12 \times 10^{-6} \) m.
- Refractive index of glass, \( \mu_g = 1.5 \).
- Wavelength of light, \( \lambda = 6000 \) Å = \( 6000 \times 10^{-10} \) m = \( 6 \times 10^{-7} \) m.
- The central bright fringe shifts a distance equal to the width of 10 bands.

2. **Calculate the Path Difference Introduced by the Glass Sheet**:
- The path difference \( \Delta x_1 \) due to the glass sheet is given by:
\[ \Delta x_1 = T_g \cdot (\mu_g - 1) \]
- Substituting the values:
\[ \Delta x_1 = 12 \times 10^{-6} \, \text{m} \cdot (1.5 - 1) = 12 \times 10^{-6} \, \text{m} \cdot 0.5 = 6 \times 10^{-6} \, \text{m} \]

3. **Relate the Path Difference to the Fringe Shift**:
- The shift of the central bright fringe is equal to the path difference caused by the glass sheet, which is equivalent to the width of 10 bands:
\[ \Delta x_1 = 10 \lambda \]
- Therefore, we can write:
\[ 10 \lambda = 6 \times 10^{-6} \, \text{m} \]

4. **Calculate the Required Path Difference for the Diamond Sheet**:
- The path difference \( \Delta x_2 \) that needs to be introduced by the diamond sheet is equal to the path difference caused by the glass sheet:
\[ \Delta x_2 = 10 \lambda \]

5. **Calculate the Thickness of the Diamond Sheet**:
- The path difference due to the diamond sheet is given by:
\[ \Delta x_2 = T_d \cdot (\mu_d - 1) \]
- Where \( \mu_d = 2.5 \) is the refractive index of diamond. Therefore:
\[ 10 \lambda = T_d \cdot (2.5 - 1) \]
- Rearranging gives:
\[ T_d = \frac{10 \lambda}{\mu_d - 1} \]
- Substituting the values:
\[ T_d = \frac{10 \cdot 6 \times 10^{-7}}{2.5 - 1} = \frac{60 \times 10^{-7}}{1.5} = 40 \times 10^{-7} \, \text{m} = 4 \times 10^{-6} \, \text{m} \]

6. **Convert to Micrometers**:
- Thus, the thickness of the diamond sheet is:
\[ T_d = 4 \, \mu m \]

### Final Answer:
The thickness of the diamond sheet required to bring the central bright fringe back to its original position is **4 micrometers**.