A two-slit inteference, experiment uses coherent light of wavelength `5 xx 10^(-7)m`. Intensity in the interference pattern for the following points are `I_(1) , I_(2) , I_(3)`, and `I_(4)`, respectively
1. A point that is close to one slit than the other by `5 xx 10^(-7)m`.
2. A point where the light waves received from the two slits are out of phase by `(4 pi)/(3)` rad.
3. A point that is closer to one slit than the other by `7.5 xx 10^(-7) m`.
4. A point where the light waves received by the two slits are out of phase by `(pi)/(36)` rad.
Then which of following statements is/are correct?

Options

A. `I_(1) gt I_(4) gt I_(2) gt I_(3)`
B. `I_(1) gt I_(4) gt I_(2) gt I_(3)`
C. `4 I_(2) = I_(4)`
D. `I_(3) = 0`

Explanation

To solve the problem step by step, we need to calculate the intensity at each specified point in the two-slit interference pattern using the provided conditions.

### Step 1: Understanding the Problem
We have a two-slit interference experiment with coherent light of wavelength \( \lambda = 5 \times 10^{-7} \, \text{m} \). We need to find the intensity at four different points based on the given conditions.

### Step 2: Formula for Intensity
The intensity at any point in the interference pattern can be calculated using the formula:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_{\text{max}} \) is the maximum intensity (intensity at the central maximum) and \( \phi \) is the phase difference.

### Step 3: Calculate Intensity at Point 1
**Condition:** A point that is closer to one slit than the other by \( 5 \times 10^{-7} \, \text{m} \).

**Path Difference:**
\[ \Delta x_1 = 5 \times 10^{-7} \, \text{m} \]

**Phase Difference:**
\[ \phi_1 = \frac{2\pi}{\lambda} \Delta x_1 = \frac{2\pi}{5 \times 10^{-7}} \times 5 \times 10^{-7} = 2\pi \]

**Intensity Calculation:**
\[ I_1 = I_{\text{max}} \cos^2\left(\frac{2\pi}{2}\right) = I_{\text{max}} \cos^2(\pi) = I_{\text{max}} \cdot 1 = I_{\text{max}} \]

### Step 4: Calculate Intensity at Point 2
**Condition:** A point where the light waves received from the two slits are out of phase by \( \frac{4\pi}{3} \, \text{rad} \).

**Intensity Calculation:**
\[ I_2 = I_{\text{max}} \cos^2\left(\frac{4\pi/3}{2}\right) = I_{\text{max}} \cos^2\left(\frac{2\pi}{3}\right) \]
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \Rightarrow \cos^2\left(\frac{2\pi}{3}\right) = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ I_2 = I_{\text{max}} \cdot \frac{1}{4} \]

### Step 5: Calculate Intensity at Point 3
**Condition:** A point that is closer to one slit than the other by \( 7.5 \times 10^{-7} \, \text{m} \).

**Path Difference:**
\[ \Delta x_3 = 7.5 \times 10^{-7} \, \text{m} \]

**Phase Difference:**
\[ \phi_3 = \frac{2\pi}{\lambda} \Delta x_3 = \frac{2\pi}{5 \times 10^{-7}} \times 7.5 \times 10^{-7} = 3\pi \]

**Intensity Calculation:**
\[ I_3 = I_{\text{max}} \cos^2\left(\frac{3\pi}{2}\right) = I_{\text{max}} \cdot 0 = 0 \]

### Step 6: Calculate Intensity at Point 4
**Condition:** A point where the light waves received by the two slits are out of phase by \( \frac{\pi}{36} \, \text{rad} \).

**Intensity Calculation:**
\[ I_4 = I_{\text{max}} \cos^2\left(\frac{\pi/36}{2}\right) = I_{\text{max}} \cos^2\left(\frac{\pi}{72}\right) \]
Using the approximation \( \cos\left(\frac{\pi}{72}\right) \approx 1 \):
\[ I_4 \approx I_{\text{max}} \]

### Step 7: Summary of Intensities
- \( I_1 = I_{\text{max}} \)
- \( I_2 = \frac{I_{\text{max}}}{4} \)
- \( I_3 = 0 \)
- \( I_4 \approx I_{\text{max}} \)

### Step 8: Comparison of Intensities
From the calculated intensities:
- \( I_1 > I_4 \)
- \( I_4 > I_2 \)
- \( I_3 = 0 < I_2 \)

### Conclusion
The correct statements based on the intensities are:
1. \( I_1 > I_4 \)
2. \( I_4 > I_2 \)
3. \( I_3 = 0 \)