To solve the problem step by step, we will follow the given information and apply the relevant equations from the Young's Double Slit Experiment (YDSE).
### Step 1: Understand the given data
- Distance between the slits and the screen, \( D = 100 \, \text{cm} = 1 \, \text{m} \)
- Initial fringe width, \( \beta_1 = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} \)
- Change in distance between the slits, \( \Delta d = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)
- New fringe width, \( \beta_2 = \frac{2}{3} \beta_1 \)
### Step 2: Write the formula for fringe width
The fringe width \( \beta \) in YDSE is given by:
\[
\beta = \frac{\lambda D}{d}
\]
where \( \lambda \) is the wavelength of light, \( D \) is the distance to the screen, and \( d \) is the distance between the slits.
### Step 3: Set up the equations for the initial and new fringe widths
For the initial fringe width:
\[
\beta_1 = \frac{\lambda D}{d} \quad \text{(1)}
\]
For the new fringe width after increasing the slit distance:
\[
\beta_2 = \frac{\lambda D}{d + \Delta d} \quad \text{(2)}
\]
Given that \( \beta_2 = \frac{2}{3} \beta_1 \), we can substitute this into equation (2):
\[
\frac{2}{3} \beta_1 = \frac{\lambda D}{d + \Delta d}
\]
### Step 4: Substitute \( \beta_1 \) from equation (1) into the new equation
Substituting \( \beta_1 \) from equation (1) into the equation for \( \beta_2 \):
\[
\frac{2}{3} \left( \frac{\lambda D}{d} \right) = \frac{\lambda D}{d + \Delta d}
\]
### Step 5: Simplify the equation
Cancelling \( \lambda D \) from both sides (assuming \( \lambda D \neq 0 \)):
\[
\frac{2}{3d} = \frac{1}{d + \Delta d}
\]
Cross-multiplying gives:
\[
2(d + \Delta d) = 3d
\]
### Step 6: Solve for \( d \)
Expanding and rearranging:
\[
2d + 2\Delta d = 3d \implies d = 2\Delta d
\]
Substituting \( \Delta d = 1.2 \times 10^{-3} \, \text{m} \):
\[
d = 2 \times 1.2 \times 10^{-3} = 2.4 \times 10^{-3} \, \text{m} = 2.4 \, \text{mm}
\]
### Step 7: Calculate the wavelength \( \lambda \)
Substituting \( d \) back into equation (1):
\[
\beta_1 = \frac{\lambda D}{d} \implies \lambda = \frac{\beta_1 d}{D}
\]
Substituting the known values:
\[
\lambda = \frac{0.25 \times 10^{-3} \times 2.4 \times 10^{-3}}{1} = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}
\]
### Step 8: Calculate the thickness of the glass plate
When a glass plate of refractive index \( \mu = 1.5 \) is placed in front of one slit, the shift in the central maximum is given as \( 20 \beta_1 \).
The formula for the shift \( S \) is:
\[
S = ( \mu - 1 ) t
\]
where \( t \) is the thickness of the plate. The shift is also given by:
\[
S = 20 \beta_1 = 20 \times 0.25 \times 10^{-3} = 5 \times 10^{-3} \, \text{m}
\]
Setting the two expressions for shift equal:
\[
(1.5 - 1) t = 5 \times 10^{-3}
\]
\[
0.5 t = 5 \times 10^{-3} \implies t = \frac{5 \times 10^{-3}}{0.5} = 10 \times 10^{-3} = 10 \, \text{mm}
\]
### Final Answers
- Wavelength \( \lambda = 600 \, \text{nm} \)
- Thickness of the glass plate \( t = 10 \, \text{mm} \)
