In the standard Young's double slit experiment, the intensity on the screen at a point distant `1.25` fringe widths from the central maximum is (assuming slits to be identical)

Options

A. `1/2 I_(max)`
B. `1/4 I_(max)`
C. `1/3 I_(max)`
D. `I_(max)`

Explanation

To solve the problem of finding the intensity on the screen at a point distant `1.25` fringe widths from the central maximum in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution

1. **Understanding Fringe Width**:
The distance `y` from the central maximum to the point of interest is given as `1.25` fringe widths. The fringe width (β) is defined as:
\[ \beta = \frac{\lambda D}{d} \]
where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

2. **Path Difference Calculation**:
The path difference (Δx) at a distance `y` from the central maximum can be expressed as:
\[ \Delta x = \frac{y \cdot d}{D} \]
Substituting \( y = 1.25 \beta \):
\[ \Delta x = 1.25 \cdot \beta = 1.25 \cdot \frac{\lambda D}{d} \]

3. **Phase Difference Calculation**:
The phase difference (φ) corresponding to the path difference is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting for Δx:
\[ \phi = \frac{2\pi}{\lambda} \cdot 1.25 \cdot \frac{\lambda D}{d} \]
Simplifying, we find:
\[ \phi = 2\pi \cdot 1.25 = 2.5\pi \]

4. **Intensity Calculation**:
The intensity (I) at a point on the screen is given by:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Substituting φ:
\[ I = I_{\text{max}} \cos^2\left(\frac{2.5\pi}{2}\right) = I_{\text{max}} \cos^2\left(1.25\pi\right) \]

5. **Evaluating Cosine**:
We know that:
\[ \cos(1.25\pi) = \cos\left(\pi + \frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \]
Therefore:
\[ \cos^2(1.25\pi) = \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

6. **Final Intensity Expression**:
Substituting back into the intensity formula:
\[ I = I_{\text{max}} \cdot \frac{1}{2} \]

### Conclusion
Thus, the intensity on the screen at a point distant `1.25` fringe widths from the central maximum is:
\[ I = \frac{I_{\text{max}}}{2} \]