Four monochromatic and coherent sources of light emitting waves in phase at placed on y axis at y = 0, a, 2a and 3a. If the intensity of wave reaching at point P far away on y axis from each of the source is almost the same and equal to `I_(0)`, then the resultant intensity at P for `a=(lambda)/(8)` is `nI_(0)`. The value of `[n]` is.
Here [] is greatest integer funciton.

Answer

To solve the problem, we need to calculate the resultant intensity at point P due to the four coherent sources of light placed on the y-axis at positions y = 0, a, 2a, and 3a. Given that the distance \( a = \frac{\lambda}{8} \), we will find the value of \( n \) in the expression for the resultant intensity \( nI_0 \).

### Step-by-Step Solution:

1. **Identify the Positions of the Sources:**
- The sources are located at:
- Source 1: \( y_1 = 0 \)
- Source 2: \( y_2 = a \)
- Source 3: \( y_3 = 2a \)
- Source 4: \( y_4 = 3a \)

2. **Calculate the Path Difference:**
- The distance between each source is \( a = \frac{\lambda}{8} \).
- The path difference between any two sources is given by the distance between them. For example, between source 1 and source 2, the path difference is \( \Delta x = a = \frac{\lambda}{8} \).

3. **Calculate the Phase Difference:**
- The phase difference \( \Delta \phi \) between two sources can be calculated using:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- For \( \Delta x = \frac{\lambda}{8} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \]

4. **Resultant Intensity Calculation:**
- The intensity at point P from each source is \( I_0 \).
- The resultant intensity \( I_R \) from two sources with intensities \( I_1 \) and \( I_2 \) and a phase difference \( \phi \) is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- For sources 1 and 3 (which are separated by \( 2a \)):
- The phase difference is \( \Delta \phi = 2 \times \frac{\pi}{4} = \frac{\pi}{2} \).
- Thus, the resultant intensity from sources 1 and 3:
\[ I_{13} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{\pi}{2}\right) = 2I_0 \]

5. **Calculate the Resultant Intensity from Sources 2 and 4:**
- Similarly, for sources 2 and 4:
- The phase difference is also \( \frac{\pi}{2} \).
- Thus, the resultant intensity from sources 2 and 4:
\[ I_{24} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{\pi}{2}\right) = 2I_0 \]

6. **Combine the Resultant Intensities:**
- Now we have two resultant intensities:
- From sources 1 and 3: \( 2I_0 \)
- From sources 2 and 4: \( 2I_0 \)
- The resultant intensity at point P is:
\[ I_R = \sqrt{(2I_0)^2 + (2I_0)^2} = \sqrt{4I_0^2 + 4I_0^2} = \sqrt{8I_0^2} = 2\sqrt{2}I_0 \]

7. **Final Resultant Intensity:**
- The total resultant intensity can be expressed as:
\[ I_R = 4I_0 + 2\sqrt{2}I_0 \]
- Using the approximation \( \sqrt{2} \approx 1.41 \):
\[ I_R \approx 4I_0 + 2 \times 1.41 I_0 = 4I_0 + 2.82I_0 = 6.82I_0 \]

8. **Determine the Value of \( n \):**
- From the problem, we have \( I_R = nI_0 \).
- Thus, \( n \approx 6.82 \).
- The greatest integer function \( [n] = 6 \).

### Final Answer:
The value of \( [n] \) is \( 6 \).