A transparent slab of thickness t and refractive index `mu` is inserted in front of upper of YDSE apparauts. The wavelength of ligth used. is `lambda`. Assume that there is no absorption of light by the slab. Mark the correct statement(s).

Options

A. The intensity of dark fringes will be 0, if slits are identical
B. The change in optical path due to insertion of plate is `mu t`.
C. The change in optical path due to insertion of plate is`(mu - 1)t`
D. For making intensity zero at center of screen, the thickness can be `(5 lambda)/(2(mu - 1))`.

Explanation

To solve the problem, we need to analyze the effect of inserting a transparent slab of thickness \( t \) and refractive index \( \mu \) in front of one of the slits in a Young's Double Slit Experiment (YDSE). We will determine the changes in optical path and the implications for interference patterns.

### Step-by-Step Solution:

1. **Understanding Optical Path**:
- The optical path length is defined as the product of the physical path length and the refractive index of the medium through which the light travels.
- In air, the optical path length is simply the distance traveled, but in a medium with refractive index \( \mu \), it is given by \( \text{Optical Path} = \text{Physical Path} \times \mu \).

2. **Path Difference Without Slab**:
- In the absence of the slab, the path difference \( \Delta x \) between the two waves arriving at point \( P \) on the screen is given by:
\[ \Delta x = S_2P - S_1P \]
- If \( P \) is at a distance \( y \) from the center, this can be expressed as:
\[ \Delta x = \frac{yD}{d} \]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

3. **Path Difference With Slab**:
- When the slab is inserted in front of slit \( S_1 \), the light traveling through \( S_1 \) will now have to pass through the slab.
- The effective path length for the wave traveling through \( S_1 \) becomes:
\[ S_1P' = S_1P - t \]
- The optical path length for the wave traveling through the slab is:
\[ \text{Optical Path for } S_1 = (S_1P - t) + \mu t = S_1P + (\mu - 1)t \]
- Therefore, the new path difference becomes:
\[ \Delta x' = S_2P - (S_1P + (\mu - 1)t) = \Delta x - (\mu - 1)t \]

4. **Condition for Dark Fringes**:
- For destructive interference (dark fringes), the path difference must be an odd multiple of half the wavelength:
\[ \Delta x' = (2n + 1) \frac{\lambda}{2}, \quad n = 0, 1, 2, \ldots \]
- Substituting for \( \Delta x' \):
\[ \Delta x - (\mu - 1)t = (2n + 1) \frac{\lambda}{2} \]

5. **Finding Thickness \( t \)**:
- Rearranging the equation gives:
\[ t = \frac{\Delta x - (2n + 1) \frac{\lambda}{2}}{\mu - 1} \]
- This shows how the thickness \( t \) relates to the path difference and the refractive index.

6. **Conclusion**:
- The correct statements regarding the effect of the slab on the interference pattern can be derived from the above analysis. Specifically, we can conclude:
- The path difference changes due to the presence of the slab.
- The conditions for dark fringes will depend on the thickness of the slab and the refractive index.