In a Young's double slit experiment , the path difference at a certain point on the screen ,between two interfering waves is `1/8`th of wavelength .The ratio of the intensity at this point to that at the center of a bright fringe is close to :

Options

A. 0.853
B. 0.568
C. 0.672
D. 0.76

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between path difference and phase difference
In a Young's double slit experiment, the path difference (Δx) and phase difference (Δφ) are related by the equation:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Given that the path difference is \( \frac{1}{8} \lambda \), we can calculate the phase difference.

### Step 2: Calculate the phase difference
Substituting the given path difference into the equation:
\[ \Delta \phi = \frac{2\pi}{\lambda} \left(\frac{1}{8} \lambda\right) = \frac{2\pi}{8} = \frac{\pi}{4} \]

### Step 3: Write the formula for intensity in terms of phase difference
The intensity \( I \) at any point on the screen in a double slit experiment can be expressed as:
\[ I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\Delta \phi) = 2I_0(1 + \cos(\Delta \phi)) \]
This can be simplified to:
\[ I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]

### Step 4: Calculate the intensity at the point with path difference \( \frac{1}{8} \lambda \)
Substituting \( \Delta \phi = \frac{\pi}{4} \) into the intensity formula:
\[ I_P = 4I_0 \cos^2\left(\frac{\pi/4}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{8}\right) \]

### Step 5: Calculate the intensity at the center of a bright fringe
At the center of a bright fringe, the phase difference is \( 0 \):
\[ I_C = 4I_0 \cos^2\left(0\right) = 4I_0 \]

### Step 6: Find the ratio of the intensity at point P to the intensity at the center of the bright fringe
Now, we can find the ratio:
\[ \frac{I_P}{I_C} = \frac{4I_0 \cos^2\left(\frac{\pi}{8}\right)}{4I_0} = \cos^2\left(\frac{\pi}{8}\right) \]

### Step 7: Calculate \( \cos^2\left(\frac{\pi}{8}\right) \)
Using a calculator or trigonometric identities, we find:
\[ \cos\left(\frac{\pi}{8}\right) \approx 0.92388 \]
Thus,
\[ \cos^2\left(\frac{\pi}{8}\right) \approx (0.92388)^2 \approx 0.85355 \]

### Final Answer
The ratio of the intensity at the point with path difference \( \frac{1}{8} \lambda \) to that at the center of a bright fringe is approximately:
\[ \frac{I_P}{I_C} \approx 0.853 \]