B

To find the intensity at the geometrical center, we need to consider the effect of the thin transparent sheet introduced in front of one of the slits.

Step 1: **Path Length Difference**
When a thin sheet of thickness t and refractive index μ is placed in front of one slit, it introduces a phase shift. The optical path length for the light passing through the sheet increases by (μ - 1)t.

Step 2: **Phase Difference Calculation**
The additional path length can be expressed in terms of phase difference. The phase difference \( \\Delta \\phi \) introduced is given by:
\( \\Delta \\phi = \frac{2\pi}{\lambda} (\mu - 1)t \).

Step 3: **Intensity Calculation**
The resultant intensity I at the center of the screen can be expressed as:
\( I = I_0 (1 + \cos(\Delta \phi)) \)
where \( I_0 \) is the intensity without the sheet.
Substituting \( \Delta \phi \):
\( I = I_0 \left(1 + \cos\left(\frac{2\pi}{\lambda} (\mu - 1)t\right)\right) \)

In our case, we assume the wave from the other slit remains unaffected. Hence the answer will depend on the values for \( \mu \) and t. Assuming ideal conditions for maximum intensity, this scenario generally leads to option B based on standard readings through similar experimental setups.