In YDSE intensity at central maxima is `I_0` The ratio `I/I_0`, at path difference `lamda/8` on the screen from central maxima , is closed to

Options

A. 0.74
B. 0.8
C. 0.9
D. 0.85

Answer

To solve the problem, we need to find the ratio of the intensity \( I \) at a path difference of \( \lambda/8 \) to the intensity at the central maximum \( I_0 \) in the Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understand the given information**:
- The intensity at the central maximum is \( I_0 \).
- The path difference at point P from the central maximum is \( \frac{\lambda}{8} \).

2. **Find the phase difference**:
- The phase difference \( \Delta \phi \) corresponding to a path difference \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting \( \Delta x = \frac{\lambda}{8} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \]

3. **Use the formula for resultant intensity**:
- The resultant intensity \( I_P \) at point P can be expressed as:
\[ I_P = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\Delta \phi) \]
- At the central maximum, \( I_1 = I_2 = I_0 \):
\[ I_P = I_0 + I_0 + 2 \sqrt{I_0 I_0} \cos\left(\frac{\pi}{4}\right) \]
- Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_P = I_0 + I_0 + 2 I_0 \cdot \frac{1}{\sqrt{2}} = 2I_0 + \frac{2I_0}{\sqrt{2}} = 2I_0 + \sqrt{2} I_0 = (2 + \sqrt{2}) I_0 \]

4. **Calculate the ratio \( \frac{I_P}{I_0} \)**:
- The ratio of the intensity at point P to the intensity at the central maximum is:
\[ \frac{I_P}{I_0} = \frac{(2 + \sqrt{2}) I_0}{I_0} = 2 + \sqrt{2} \]

5. **Approximate the value**:
- Since \( \sqrt{2} \approx 1.414 \):
\[ 2 + \sqrt{2} \approx 2 + 1.414 = 3.414 \]

6. **Find the ratio \( \frac{I}{I_0} \)**:
- The intensity at the central maximum \( I_0 \) is 4 times the intensity at the maximum, so:
\[ I_{max} = 4I_0 \]
- Therefore, the ratio \( \frac{I_P}{I_{max}} \) is:
\[ \frac{I_P}{I_{max}} = \frac{(2 + \sqrt{2}) I_0}{4 I_0} = \frac{2 + \sqrt{2}}{4} \]
- Simplifying gives:
\[ \frac{I_P}{I_{max}} = \frac{2 + 1.414}{4} \approx \frac{3.414}{4} \approx 0.8535 \]

### Final Result:
The ratio \( \frac{I}{I_0} \) at a path difference of \( \frac{\lambda}{8} \) is approximately **0.853**.