In YDSE path difference at a point on screen is `lambda/8` . Find ratio of intensity at this point with maximum intensity.

Options

A. `0.853`
B. `0.533`
C. `0.234`
D. `0.123`

Explanation

To solve the problem, we need to find the ratio of the intensity at a point on the screen in Young's Double Slit Experiment (YDSE) where the path difference is \( \frac{\lambda}{8} \) to the maximum intensity.

### Step-by-Step Solution:

1. **Understanding the Maximum Intensity**:
In YDSE, the maximum intensity (\( I_{\text{max}} \)) occurs when the two waves are in phase. If we denote the intensity of each wave as \( I_0 \), then the maximum intensity can be calculated as:
\[ I_{\text{max}} = (I_1 + I_2 + 2\sqrt{I_1 I_2}) \text{ where } I_1 = I_2 = I_0 \]
Therefore,
\[ I_{\text{max}} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 2I_0 + 2I_0 = 4I_0 \]

2. **Calculating the Phase Difference**:
The path difference (\( \Delta x \)) is given as \( \frac{\lambda}{8} \). The phase difference (\( \phi \)) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{\lambda}{8} \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \]

3. **Calculating the Intensity at Point P**:
The intensity at point P (\( I_P \)) where the phase difference is \( \frac{\pi}{4} \) can be calculated as:
\[ I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
Since \( I_1 = I_2 = I_0 \):
\[ I_P = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{\pi}{4}\right) \]
Knowing that \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_P = 2I_0 + 2I_0 \cdot \frac{1}{\sqrt{2}} = 2I_0 + \frac{2I_0}{\sqrt{2}} = 2I_0 \left(1 + \frac{1}{\sqrt{2}}\right) \]

4. **Finding the Ratio of Intensities**:
Now we need to find the ratio of the intensity at point P to the maximum intensity:
\[ \text{Ratio} = \frac{I_P}{I_{\text{max}}} = \frac{2I_0 \left(1 + \frac{1}{\sqrt{2}}\right)}{4I_0} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{1 + \frac{1}{\sqrt{2}}}{2} \]

5. **Final Calculation**:
To simplify further:
\[ \text{Ratio} = \frac{1 + \frac{1}{\sqrt{2}}}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}} = \frac{1.414 + 1}{2 \cdot 1.414} \approx \frac{2.414}{2.828} \approx 0.853 \]

### Final Answer:
The ratio of intensity at this point to the maximum intensity is approximately \( 0.853 \).