To solve the problem step by step, we will analyze the given information and apply the principles of wave optics.
### Given Data:
- Thickness of each slab, \( t = 0.45 \, \text{mm} = 0.45 \times 10^{-3} \, \text{m} \)
- Refractive indices: \( n_1 = 1.40 \), \( n_2 = 1.42 \)
- Separation of slits, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance from slits to screen, \( L = 1 \, \text{m} \)
### Step 1: Calculate the Optical Path Length Difference
The optical path length for a medium is given by:
\[
\text{Optical Path Length} = n \cdot t
\]
For the first slab (with \( n_1 = 1.40 \)):
\[
\text{OPL}_1 = n_1 \cdot t = 1.40 \cdot 0.45 \times 10^{-3} \, \text{m} = 0.000630 \, \text{m}
\]
For the second slab (with \( n_2 = 1.42 \)):
\[
\text{OPL}_2 = n_2 \cdot t = 1.42 \cdot 0.45 \times 10^{-3} \, \text{m} = 0.000639 \, \text{m}
\]
### Step 2: Calculate the Path Difference
The path difference (\( \Delta x \)) due to the slabs is:
\[
\Delta x = \text{OPL}_2 - \text{OPL}_1 = 0.000639 \, \text{m} - 0.000630 \, \text{m} = 0.000009 \, \text{m} = 9 \, \mu\text{m}
\]
### Step 3: Relate Path Difference to Position on the Screen
The path difference can also be expressed in terms of the position \( y \) on the screen:
\[
\Delta x = \frac{d \cdot y}{L}
\]
Setting the two expressions for path difference equal gives:
\[
\frac{d \cdot y}{L} = 9 \times 10^{-6} \, \text{m}
\]
### Step 4: Solve for \( y \)
Substituting the known values:
\[
\frac{1 \times 10^{-3} \cdot y}{1} = 9 \times 10^{-6}
\]
\[
y = \frac{9 \times 10^{-6}}{1 \times 10^{-3}} = 0.009 \, \text{m} = 9 \, \text{mm}
\]
### Conclusion
The position of the central maxima from the center of the screen is:
\[
\boxed{9 \, \text{mm}}
\]
