In `YDSE`, slab of thickness t and refractive index `mu` is placed in front of any slit. Then displacement of central maximu is terms of fringe width when light of wavelength `lamda` is incident on system is

Options

A. `(beta(mu-1)t)/(2lamda)`
B. `(beta(mu-1)t)/(lamda)`
C. `(beta(mu-1)t)/(3lamda)`
D. `(beta(mu-1)t)/(4lamda)`

Answer

To solve the problem of finding the displacement of the central maximum in Young's Double Slit Experiment (YDSE) when a slab of thickness \( t \) and refractive index \( \mu \) is placed in front of one of the slits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In YDSE, we have two slits and a screen where interference patterns are formed. When a slab is placed in front of one of the slits, it changes the optical path length for the light passing through that slit.

2. **Calculate the Optical Path Difference**:
- The optical path length through the slab is given by:
\[ \text{Optical Path Length} = \mu t \]
- The physical path length without the slab would just be \( t \). Therefore, the optical path difference introduced by the slab is:
\[ \Delta x = \mu t - t = t(\mu - 1) \]

3. **Relating Optical Path Difference to Displacement**:
- The displacement of the central maximum \( y \) on the screen can be related to the path difference. The condition for constructive interference (central maximum) is that the path difference should equal an integer multiple of the wavelength \( \lambda \).
- For small angles, the physical path difference can be expressed as:
\[ \Delta x = y \frac{d}{D} \]
- Here, \( d \) is the distance between the slits, \( D \) is the distance from the slits to the screen, and \( y \) is the displacement of the central maximum.

4. **Setting the Path Differences Equal**:
- Setting the optical path difference equal to the physical path difference gives:
\[ t(\mu - 1) = y \frac{d}{D} \]

5. **Solving for Displacement \( y \)**:
- Rearranging the equation to solve for \( y \):
\[ y = \frac{t(\mu - 1) D}{d} \]

6. **Expressing in Terms of Fringe Width**:
- The fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
- Substituting this into our expression for \( y \):
\[ y = \frac{t(\mu - 1)}{\lambda} \cdot \beta \]

7. **Final Expression**:
- Thus, the displacement of the central maximum in terms of the fringe width \( \beta \) is:
\[ y = \beta \frac{t(\mu - 1)}{\lambda} \]

### Conclusion:
The displacement of the central maximum due to the slab is given by:
\[ y = \beta \frac{t(\mu - 1)}{\lambda} \]