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

Options

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

Answer

To solve the problem, we will use the formula for intensity in the Young's double slit experiment, which is given by:

\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]

Where:
- \( I \) is the total intensity at a point on the screen.
- \( I_1 \) and \( I_2 \) are the intensities of the light from each slit.
- \( \phi \) is the phase difference between the two waves arriving at the point.

### Step 1: Determine the intensity at path difference \( \lambda \)

Given that the intensity at a point where the path difference is \( \lambda \) is \( K \), we can express this as:

- When the path difference is \( \lambda \), the phase difference \( \phi \) is \( 2\pi \) (since \( \phi = \frac{2\pi}{\lambda} \times \text{path difference} \)).
- The intensity at this point can be expressed as:

\[ K = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(2\pi) \]

Since \( \cos(2\pi) = 1 \), we have:

\[ K = I_1 + I_2 + 2\sqrt{I_1 I_2} \]

Assuming \( I_1 = I_2 = I_0 \), we can rewrite this as:

\[ K = 4I_0 \]

### Step 2: Determine the intensity at path difference \( \frac{\lambda}{4} \)

Now, we need to find the intensity at a point where the path difference is \( \frac{\lambda}{4} \):

- The phase difference \( \phi \) at this point is:

\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]

Now, substituting this into the intensity formula:

\[ I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\left(\frac{\pi}{2}\right) \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:

\[ I' = I_1 + I_2 \]

Again, assuming \( I_1 = I_2 = I_0 \):

\[ I' = I_0 + I_0 = 2I_0 \]

### Step 3: Relate \( I_0 \) to \( K \)

From our earlier calculation, we found that:

\[ K = 4I_0 \implies I_0 = \frac{K}{4} \]

### Step 4: Substitute \( I_0 \) back into the intensity equation

Now substituting \( I_0 \) into the expression for \( I' \):

\[ I' = 2I_0 = 2 \left(\frac{K}{4}\right) = \frac{K}{2} \]

### Conclusion

Thus, the intensity at a point where the path difference is \( \frac{\lambda}{4} \) is:

\[ \boxed{\frac{K}{2}} \]