In Young's experiment intensity at a point on the scrren is 75% of the maximum value. Minimum phase difference between the waves arriving at this point from the two slits will be

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `135^(@)`

Explanation

To find the minimum phase difference between the waves arriving at a point on the screen in Young's experiment where the intensity is 75% of the maximum value, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The intensity at point P on the screen is given as 75% of the maximum intensity (I₀).
- This can be expressed as:
\[ I_P = 0.75 I_0 = \frac{3}{4} I_0 \]

2. **Using the Intensity Formula:**
- The intensity at any point in Young's double-slit experiment can be expressed in terms of the phase difference (Δφ) between the two waves as:
\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
- Therefore, for point P, we can write:
\[ I_P = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]

3. **Setting Up the Equation:**
- Substitute the expression for \(I_P\) into the intensity formula:
\[ \frac{3}{4} I_0 = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
- Dividing both sides by \(I_0\) (assuming \(I_0 \neq 0\)):
\[ \frac{3}{4} = \cos^2\left(\frac{\Delta \phi}{2}\right) \]

4. **Solving for the Cosine:**
- Taking the square root of both sides gives:
\[ \cos\left(\frac{\Delta \phi}{2}\right) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

5. **Finding the Phase Difference:**
- The cosine value of \(\frac{\sqrt{3}}{2}\) corresponds to an angle of:
\[ \frac{\Delta \phi}{2} = \frac{\pi}{6} \quad \text{(or 30 degrees)} \]
- Therefore, multiplying by 2 gives:
\[ \Delta \phi = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \quad \text{(or 60 degrees)} \]

### Final Answer:
The minimum phase difference between the waves arriving at this point from the two slits is:
\[ \Delta \phi = \frac{\pi}{3} \text{ radians} \quad \text{(or 60 degrees)} \]