Minimum thickness of a mica sheet having `mu=(3)/(2)` which should be placed in front of one of the slits in YDSE is required to reduce the intensity at the centre of screen to half of maximum intensity is

Options

A. `lamda/4`
B. `lamda/8`
C. `lamda/2`
D. `lamda/3`

Answer

To solve the problem of finding the minimum thickness of a mica sheet required to reduce the intensity at the center of the screen in a Young's Double Slit Experiment (YDSE) to half of the maximum intensity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Maximum Intensity**:
In YDSE, the maximum intensity \( I_{\text{max}} \) at the center of the screen is given by:
\[ I_{\text{max}} = 4I \]
where \( I \) is the intensity from each slit.

2. **Finding Half of Maximum Intensity**:
To find the intensity that is half of the maximum intensity:
\[ I_{\text{half}} = \frac{I_{\text{max}}}{2} = \frac{4I}{2} = 2I \]

3. **Phase Difference Calculation**:
When a mica sheet of thickness \( t \) and refractive index \( \mu = \frac{3}{2} \) is placed in front of one of the slits, it introduces a path difference \( \Delta x \):
\[ \Delta x = (\mu - 1) t = \left(\frac{3}{2} - 1\right) t = \frac{1}{2} t \]

4. **Phase Difference Relation**:
The phase difference \( \phi \) due to this path difference is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{1}{2} t\right) = \frac{\pi t}{\lambda} \]

5. **Setting Up the Intensity Equation**:
The net intensity \( I_{\text{net}} \) at the center of the screen can be expressed as:
\[ I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
Since \( I_1 = I_2 = I \):
\[ I_{\text{net}} = I + I + 2I \cos \phi = 2I + 2I \cos \phi = 2I(1 + \cos \phi) \]

6. **Setting the Intensity to Half Maximum**:
We want \( I_{\text{net}} = 2I \):
\[ 2I(1 + \cos \phi) = 2I \]
Dividing both sides by \( 2I \) (assuming \( I \neq 0 \)):
\[ 1 + \cos \phi = 1 \implies \cos \phi = 0 \]

7. **Finding the Corresponding Phase Difference**:
The condition \( \cos \phi = 0 \) occurs when:
\[ \phi = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For minimum thickness, we take \( n = 0 \):
\[ \phi = \frac{\pi}{2} \]

8. **Substituting Phase Difference into the Equation**:
Now substituting \( \phi = \frac{\pi}{2} \) into the phase difference equation:
\[ \frac{\pi t}{\lambda} = \frac{\pi}{2} \]

9. **Solving for Thickness**:
Canceling \( \pi \) from both sides:
\[ \frac{t}{\lambda} = \frac{1}{2} \implies t = \frac{\lambda}{2} \]

### Final Answer:
The minimum thickness of the mica sheet required is:
\[ t = \frac{\lambda}{2} \]