In YDSE, the sources is place symmetrical to the slits. If a transparent slab is placed in front of the upper slit, then

Options

A. intensity of central maxima may change
B. intensity of central maxima may not change
C. central maxima will be shifted up
D. intensity of dark fringe will be always zero

Explanation

To solve the problem regarding Young's Double Slit Experiment (YDSE) with a transparent slab placed in front of the upper slit, we can analyze the effects step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In YDSE, we have two slits (let's call them S1 and S2) and a coherent light source placed symmetrically with respect to these slits.
- The interference pattern is formed on a screen due to the superposition of light waves coming from both slits.

2. **Introducing the Transparent Slab**:
- A transparent slab is placed in front of the upper slit (S1). This slab has a refractive index \( \mu \) greater than that of air (which is approximately 1).
- The slab will introduce a phase shift in the light wave passing through it.

3. **Effect on Intensity of Central Maxima**:
- The intensity of the central maxima is determined by the amplitudes of the waves coming from both slits.
- When the slab is placed in front of S1, some light may be reflected and some transmitted. The transmitted light will have its amplitude reduced due to the slab.
- Therefore, the intensity of the central maxima will likely decrease because the effective amplitude of the wave from S1 is reduced.

4. **Possibility of No Change in Intensity**:
- It is also possible that if the slab is perfectly transparent and does not absorb any light, the intensity of the central maxima may not change significantly.
- Thus, the intensity may remain the same if the slab does not affect the amplitude of the transmitted light.

5. **Shift of Central Maxima**:
- The introduction of the slab causes a phase change in the light wave passing through it. The phase change \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} (t(\mu - 1)) \]
where \( t \) is the thickness of the slab and \( \lambda \) is the wavelength of light.
- Since the slab is in front of S1, the central maximum will shift upwards (towards the slit with the slab) due to the additional optical path length introduced by the slab.

6. **Intensity of Dark Fringes**:
- The dark fringes occur where the path difference between the two waves is an odd multiple of half wavelengths.
- If the intensity of the light from one slit is reduced due to the slab, the condition for dark fringes will change, and the minimum intensity may not be exactly zero anymore. Hence, the intensity of dark fringes may be greater than zero.

### Conclusion:
Based on the analysis, we can conclude:
- The intensity of the central maxima may decrease (Option A is correct).
- The intensity of the central maxima may not change (Option B is also correct).
- The central maxima will shift upwards (Option C is correct).
- The intensity of the dark fringes will not always be zero (Option D is incorrect).

### Final Answer:
The correct options are A, B, and C.