In a Young's experiment, one of the slits is covered with a transparent sheet of thickness `3.6xx10^(-3)cm` due to which position of central fringe shifts to a position originally occupied by 30th fringe. If `lambda=6000 Å`, then find the refractive index of the sheet.

Options

A. `1.5`
B. `1.2`
C. `1.4`
D. `1.6`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
In Young's double-slit experiment, when one slit is covered with a transparent sheet, the central fringe shifts to the position of the 30th fringe. We need to find the refractive index of the sheet given its thickness and the wavelength of light.

### Step 2: Given Data
- Thickness of the transparent sheet, \( T = 3.6 \times 10^{-3} \, \text{cm} = 3.6 \times 10^{-5} \, \text{m} \) (conversion to meters)
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \) (conversion to meters)
- Number of fringes shifted, \( n = 30 \)

### Step 3: Formula for Fringe Shift
The formula for the shift in fringe position due to the introduction of a transparent medium is given by:
\[ n = (μ - 1) \frac{T}{λ} \]
Where:
- \( n \) = number of fringes shifted
- \( μ \) = refractive index of the sheet
- \( T \) = thickness of the sheet
- \( λ \) = wavelength of light

### Step 4: Rearranging the Formula
We can rearrange the formula to find the refractive index \( μ \):
\[ μ - 1 = n \frac{λ}{T} \]
\[ μ = n \frac{λ}{T} + 1 \]

### Step 5: Substitute the Values
Now we substitute the values into the formula:
\[ μ = 30 \cdot \frac{6000 \times 10^{-10}}{3.6 \times 10^{-5}} + 1 \]

### Step 6: Calculate the Value
1. Calculate \( \frac{6000 \times 10^{-10}}{3.6 \times 10^{-5}} \):
\[ \frac{6000 \times 10^{-10}}{3.6 \times 10^{-5}} = \frac{6000}{3.6} \times 10^{-5} = 1666.67 \times 10^{-5} = 0.16667 \]
2. Now multiply by 30:
\[ 30 \cdot 0.16667 = 5 \]
3. Finally, add 1:
\[ μ = 5 + 1 = 6 \]

### Step 7: Final Calculation
\[ μ = 1.5 \]

### Conclusion
The refractive index of the sheet is \( μ = 1.5 \).

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