Three waves from three coherent sources meet at some point. Resultant amplitude of each is `A_0`. Intensity corresponding to `A_0 is I_0`. Phase difference between first wave and second wave is `60^@`. Path difference between first wave and third wave is `lambda/3`. The first wave lags behind in phase angle from second and third wave. Find resultant intensity at this point.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have three coherent waves with the following properties:
- Amplitude of each wave: \( A_0 \)
- Intensity corresponding to \( A_0 \): \( I_0 \)
- Phase difference between the first wave and the second wave: \( 60^\circ \)
- Path difference between the first wave and the third wave: \( \frac{\lambda}{3} \)

### Step 2: Calculate the Phase Difference Between the First and Third Waves
The path difference of \( \frac{\lambda}{3} \) corresponds to a phase difference given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \times \text{path difference} = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \]
Converting this to degrees:
\[ \Delta \phi = \frac{2\pi}{3} \times \frac{180^\circ}{\pi} = 120^\circ \]

### Step 3: Represent the Waves in Phasor Form
Let:
- Wave 1 (first wave) be represented as \( A_0 e^{i0} \) (reference wave)
- Wave 2 (second wave) be represented as \( A_0 e^{i60^\circ} \)
- Wave 3 (third wave) be represented as \( A_0 e^{i120^\circ} \)

### Step 4: Calculate the Resultant Amplitude of Waves 1 and 3
To find the resultant amplitude of waves 1 and 3, we can use the formula for the resultant of two phasors:
\[ R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
Where:
- \( A_1 = A_0 \)
- \( A_2 = A_0 \)
- \( \phi = 120^\circ \)

Substituting the values:
\[ R = \sqrt{A_0^2 + A_0^2 + 2 A_0 A_0 \cos(120^\circ)} \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ R = \sqrt{A_0^2 + A_0^2 - A_0^2} = \sqrt{A_0^2} = A_0 \]

### Step 5: Combine the Resultant Amplitude with Wave 2
Now, we need to find the resultant amplitude when we add this resultant amplitude \( R \) from waves 1 and 3 to wave 2:
\[ R_{\text{total}} = R + A_0 e^{i60^\circ} \]
Since \( R = A_0 \):
\[ R_{\text{total}} = A_0 + A_0 e^{i60^\circ} \]

### Step 6: Calculate the Resultant Amplitude
Using the cosine rule:
\[ R_{\text{total}} = A_0 \left(1 + e^{i60^\circ}\right) \]
The magnitude of \( e^{i60^\circ} = \cos(60^\circ) + i\sin(60^\circ) = \frac{1}{2} + i\frac{\sqrt{3}}{2} \):
\[ R_{\text{total}} = A_0 \left(1 + \frac{1}{2}\right) = A_0 \cdot \frac{3}{2} = \frac{3A_0}{2} \]

### Step 7: Calculate the Resultant Intensity
The intensity is proportional to the square of the amplitude:
\[ I = k R_{\text{total}}^2 \]
Where \( k \) is a proportionality constant. Since \( I_0 \) corresponds to \( A_0^2 \):
\[ I = k \left(\frac{3A_0}{2}\right)^2 = k \cdot \frac{9A_0^2}{4} \]
Thus, the resultant intensity is:
\[ I = \frac{9}{4} I_0 \]

### Final Answer
The resultant intensity at the point where the three waves meet is:
\[ \boxed{\frac{9}{4} I_0} \]