In a YDSE with two identical slits, when the upper slit is covered with a thin, perfectly tranparent sheet of mica, the intensity at the centre of screen reduces ro `75%` of the initial value. The second minima is observed to above this point and third maxima below it. which of the following can not be a possible value of phase difference caused by the mica sheet

Options

A. `(pi)/(3)`
B. `(7pi)/(2)`
C. `(10pi)/(3)`
D. `(11pi)/(3)`

Answer

To solve the problem, we need to analyze the situation described in the Young's Double Slit Experiment (YDSE) after a mica sheet is placed over one of the slits. The key points to consider are the intensity reduction and the conditions for minima and maxima.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- Let the initial intensity at the center of the screen (without the mica sheet) be \( I_0 \).
- When the mica sheet is placed, the intensity reduces to \( I' = 0.75 I_0 \).

2. **Intensity Formula**:
- The intensity in a YDSE is given by:
\[ I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]
- Here, \( \phi \) is the phase difference introduced by the mica sheet.

3. **Setting Up the Equation**:
- For the new intensity after placing the mica sheet:
\[ 0.75 I_0 = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]
- Dividing both sides by \( I_0 \):
\[ 0.75 = 4 \cos^2\left(\frac{\phi}{2}\right) \]
- Rearranging gives:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{0.75}{4} = \frac{3}{16} \]

4. **Finding Phase Difference**:
- Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{4} \]
- Therefore, the angle corresponding to this cosine value is:
\[ \frac{\phi}{2} = \cos^{-1}\left(\frac{\sqrt{3}}{4}\right) \]
- This gives two possible values for \( \frac{\phi}{2} \): \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \).

5. **Calculating Possible Values of \( \phi \)**:
- Thus, the possible values of \( \phi \) are:
\[ \phi = \frac{\pi}{3}, \quad \frac{5\pi}{3} \]
- Since the phase difference can be expressed in multiples of \( 2\pi \), we can also consider:
\[ \phi = 2n\pi + \frac{\pi}{3} \quad \text{or} \quad \phi = 2n\pi + \frac{5\pi}{3} \quad (n \in \mathbb{Z}) \]

6. **Identifying Minima and Maxima**:
- The second minima occurs at:
\[ \phi = (2n-1)\pi \quad \text{for } n=2 \Rightarrow \phi = 3\pi \]
- The third maxima occurs at:
\[ \phi = 2n\pi \quad \text{for } n=3 \Rightarrow \phi = 6\pi \]

7. **Conclusion**:
- The phase difference \( \phi \) must lie between \( 3\pi \) and \( 6\pi \).
- Possible values of \( \phi \) include \( \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3} \), etc.
- Values like \( \frac{\pi}{3} \) do not lie in this range and thus cannot be a possible value of the phase difference caused by the mica sheet.

### Final Answer:
The value that cannot be a possible value of phase difference caused by the mica sheet is \( \frac{\pi}{3} \).