In Young’s double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is `frac{7 lambda}{4}`. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

Options

A. `1/2`
B. `3/4`
C. `1/3`
D. `1/4`

Answer

To solve the problem, we need to find the ratio of the intensity of the fringe at a point on the screen (where the path difference is given) to the maximum intensity of the fringe in Young's double slit experiment.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Path difference \( \Delta x = \frac{7\lambda}{4} \)
- Both sources have the same intensity \( I_1 = I_2 = I \)

2. **Calculate the Phase Difference:**
The phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the given path difference:
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{7\lambda}{4} = \frac{7\pi}{2} \]

3. **Use the Resultant Intensity Formula:**
The resultant intensity \( I_R \) is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \]
Substituting \( I_1 = I \) and \( I_2 = I \):
\[ I_R = I + I + 2\sqrt{I \cdot I} \cos\left(\frac{7\pi}{2}\right) \]
Simplifying this:
\[ I_R = 2I + 2I \cos\left(\frac{7\pi}{2}\right) \]

4. **Evaluate \( \cos\left(\frac{7\pi}{2}\right) \):**
The angle \( \frac{7\pi}{2} \) can be simplified. Since \( \cos\) has a periodicity of \( 2\pi \):
\[ \frac{7\pi}{2} = 3\pi + \frac{\pi}{2} = 2\pi + \pi + \frac{\pi}{2} \implies \cos\left(\frac{7\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

5. **Substituting Back into the Intensity Formula:**
Now substituting \( \cos\left(\frac{7\pi}{2}\right) = 0 \) into the intensity formula:
\[ I_R = 2I + 2I \cdot 0 = 2I \]

6. **Calculate the Maximum Intensity:**
The maximum intensity \( I_{max} \) occurs when the phase difference is \( 0 \):
\[ I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cdot 1 = I + I + 2I = 4I \]

7. **Find the Ratio of Intensities:**
Now, we can find the ratio of the resultant intensity to the maximum intensity:
\[ \text{Ratio} = \frac{I_R}{I_{max}} = \frac{2I}{4I} = \frac{1}{2} \]

### Final Answer:
The ratio of the intensity of the fringe at the given point to the maximum intensity is \( \frac{1}{2} \).