In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is `1/3 rd` of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

Options

A. `0.25`
B. `0.67`
C. `0.75`
D. `0.56`

Explanation

To solve the problem, we need to find the ratio of the intensity of light at a point on the screen where the path difference is \( \frac{1}{3} \) of a wavelength to the intensity at the center of a bright fringe in a double-slit experiment.

### Step-by-Step Solution:

1. **Understanding Path Difference**:
The path difference \( \Delta x \) between the two waves at the point of interest is given as:
\[ \Delta x = \frac{\lambda}{3} \]

2. **Calculating Phase Difference**:
The phase difference \( \Delta \phi \) corresponding to the path difference can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{\lambda}{3} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \]

3. **Intensity Calculation**:
The intensity \( I \) at any point in a double-slit experiment can be expressed in terms of the intensity of individual waves \( I_0 \) and the phase difference:
\[ I = I_0 \left(1 + \cos \Delta \phi\right) \]
Using \( \Delta \phi = \frac{2\pi}{3} \):
\[ I = I_0 \left(1 + \cos \frac{2\pi}{3}\right) \]
We know that \( \cos \frac{2\pi}{3} = -\frac{1}{2} \):
\[ I = I_0 \left(1 - \frac{1}{2}\right) = I_0 \cdot \frac{1}{2} \]

4. **Maximum Intensity**:
The maximum intensity \( I_{\text{max}} \) at the center of a bright fringe occurs when the phase difference is zero (constructive interference):
\[ I_{\text{max}} = 4I_0 \]
(This is because the intensity is proportional to the square of the amplitude, and for two waves, \( I_{\text{max}} = 4I_0 \) when they are in phase.)

5. **Finding the Ratio**:
Now, we find the ratio of the intensity at the point with path difference \( \frac{\lambda}{3} \) to the maximum intensity:
\[ \text{Ratio} = \frac{I}{I_{\text{max}}} = \frac{I_0 \cdot \frac{1}{2}}{4I_0} = \frac{1/2}{4} = \frac{1}{8} \]

### Final Answer:
The ratio of the intensity of light at that point to that at the center of a bright fringe is:
\[ \frac{1}{8} \]