In YDSE, when a glass plate of refractive `1.5` of thickness `t` is placed in the path of one of the interfering beams of wavelength `lamda`, intensity at the position where central maximum occurred previously remain unchanged. If the minimum thicknes of the glass plate is `klamda`. Find the value of `k`........

Explanation

To solve the problem, we need to determine the value of \( k \) given that the minimum thickness of the glass plate is \( k \lambda \) and it does not change the intensity at the position of the central maximum in Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understanding the Shift in Central Maximum**:
In YDSE, when a glass plate is introduced in one of the beams, it causes a phase shift. The phase shift \( \Delta \phi \) due to the glass plate can be calculated using the formula:
\[ \Delta \phi = \frac{2 \pi}{\lambda} \times \Delta x \]
where \( \Delta x \) is the optical path difference caused by the glass plate.

2. **Calculating the Optical Path Difference**:
The optical path difference introduced by a glass plate of thickness \( t \) and refractive index \( \mu \) is given by:
\[ \Delta x = (\mu - 1) t \]
Here, \( \mu = 1.5 \) for the glass plate.

3. **Substituting Values**:
Substituting the value of \( \mu \):
\[ \Delta x = (1.5 - 1) t = 0.5 t \]

4. **Condition for Unchanged Intensity**:
For the intensity at the central maximum to remain unchanged, the phase shift must be an integral multiple of \( 2\pi \):
\[ \Delta \phi = 2n\pi \quad (n = 0, 1, 2, \ldots) \]
Therefore, we can write:
\[ \frac{2 \pi}{\lambda} \times 0.5 t = 2n\pi \]
Simplifying this gives:
\[ 0.5 t = 2n\lambda \]

5. **Finding Minimum Thickness**:
Rearranging for \( t \):
\[ t = 4n\lambda \]
The minimum thickness occurs when \( n = 1 \):
\[ t_{min} = 4 \lambda \]

6. **Relating Thickness to \( k \)**:
Given that the minimum thickness of the glass plate is \( k \lambda \):
\[ k \lambda = 4 \lambda \]
Therefore, we find:
\[ k = 4 \]

### Final Answer:
The value of \( k \) is \( 4 \).

---