In Young`s double slit experiment, the phase different between two coherent sources of equal intensity is π/3.The intensity at a point which is at equal distance from the two slits is (l0 is maximum intensity)

Options

A. (`l_@)/2`
B. (`3l_@)/4`
C. (`l_@)/4`
D. `(l_@)/sqrt2`

Answer

To solve the problem of finding the intensity at a point in Young's double slit experiment where the phase difference between two coherent sources is π/3, we can follow these steps:

### Step 1: Understand the given information
We know that in Young's double slit experiment:
- The phase difference (Δφ) between the two coherent sources (S1 and S2) is given as π/3.
- The maximum intensity (I0) is the intensity at the point where constructive interference occurs.

### Step 2: Use the formula for intensity
The intensity at a point (Ip) in terms of the maximum intensity (I0) and the phase difference (Δφ) can be expressed as:
\[ I_p = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]

### Step 3: Substitute the phase difference
Substituting the given phase difference (Δφ = π/3) into the formula:
\[ I_p = I_0 \cos^2\left(\frac{\pi}{3}/2\right) \]
This simplifies to:
\[ I_p = I_0 \cos^2\left(\frac{\pi}{6}\right) \]

### Step 4: Calculate cos(π/6)
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus:
\[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 5: Substitute back to find Ip
Now substituting this value back into the intensity formula:
\[ I_p = I_0 \cdot \frac{3}{4} \]

### Final Answer
Thus, the intensity at the point is:
\[ I_p = \frac{3}{4} I_0 \]