In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is `lambda` is K, (`lambda` being the wavelength of light used). The intensity at a point where the path difference is `lambda//4` will be

Options

A. K
B. `K//4`
C. `K//2`
D. Zero

Explanation

To solve the problem, we will use the principles of interference and the relationship between intensity and phase difference in the Young's double-slit experiment.

### Step-by-Step Solution:

1. **Understanding the Intensity Formula**:
In Young's double-slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the intensity of the individual slits \( I_0 \) and the phase difference \( \phi \) between the waves arriving at that point:
\[ I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]

2. **Finding Intensity at Path Difference \( \lambda \)**:
Given that the intensity at a point where the path difference is \( \lambda \) is \( K \), we can find the phase difference \( \phi \) corresponding to this path difference:
\[ \phi = \frac{2\pi}{\lambda} \cdot \text{(path difference)} = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]
Substituting this into the intensity formula:
\[ K = 4 I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4 I_0 \cos^2(\pi) = 4 I_0 \cdot 1 = 4 I_0 \]

3. **Finding Intensity at Path Difference \( \frac{\lambda}{4} \)**:
Now, we need to find the intensity at a point where the path difference is \( \frac{\lambda}{4} \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]
Now substituting this phase difference into the intensity formula:
\[ I = 4 I_0 \cos^2\left(\frac{\pi}{2}\right) = 4 I_0 \cdot 0 = 0 \]

4. **Relating Intensities**:
Since we have already established that \( K = 4 I_0 \), we can express \( I \) in terms of \( K \):
\[ I = 0 \implies \text{Intensity at path difference } \frac{\lambda}{4} = 0 \]

### Conclusion:
The intensity at a point where the path difference is \( \frac{\lambda}{4} \) is \( 0 \).