In YDSE, a glass slab of refractive index, `mu= 1.5` and thickness 'l' is introduced in one of the interfening beams of wavelength `lambda = 5000 A`. If on introducing the slab, the central fringe shift by 2 mm, then thickness of stab would be (Fringe width, `beta = 0.2 mm)`

Options

A. `3*10^-3 mm`
B. `1*10^-4 mm`
C. `1*10^-2 mm`
D. `2*10^-3 mm`

Answer

To solve the problem, we will use the formula that relates the fringe shift (S) to the thickness (t) of the glass slab, the wavelength (λ), and the refractive index (μ). The formula is given by:

\[ S = \frac{\beta}{\lambda} \times (μ - 1) \times t \]

Where:
- \( S \) = fringe shift
- \( \beta \) = fringe width
- \( λ \) = wavelength of light
- \( μ \) = refractive index of the glass slab
- \( t \) = thickness of the glass slab

### Step-by-Step Solution:

1. **Convert the given values to consistent units**:
- Wavelength \( λ = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5000 \times 10^{-7} \, \text{mm} = 5 \times 10^{-4} \, \text{mm} \)
- Fringe shift \( S = 2 \, \text{mm} \)
- Fringe width \( β = 0.2 \, \text{mm} \)
- Refractive index \( μ = 1.5 \)

2. **Insert the values into the formula**:
\[ S = \frac{\beta}{\lambda} \times (μ - 1) \times t \]
Plugging in the values:
\[ 2 = \frac{0.2}{5 \times 10^{-4}} \times (1.5 - 1) \times t \]

3. **Calculate \( (μ - 1) \)**:
\[ μ - 1 = 1.5 - 1 = 0.5 \]

4. **Rearranging the equation to solve for \( t \)**:
\[ 2 = \frac{0.2}{5 \times 10^{-4}} \times 0.5 \times t \]
\[ t = \frac{2 \times 5 \times 10^{-4}}{0.2 \times 0.5} \]

5. **Calculate the right side**:
- Calculate the numerator:
\[ 2 \times 5 \times 10^{-4} = 10 \times 10^{-4} = 1 \times 10^{-3} \, \text{mm} \]
- Calculate the denominator:
\[ 0.2 \times 0.5 = 0.1 \]
- Now, substituting back:
\[ t = \frac{1 \times 10^{-3}}{0.1} = 10 \times 10^{-3} = 1 \, \text{mm} \]

### Final Answer:
The thickness of the glass slab \( t \) is **1 mm**.