In Young's double-slit experiment using monochromatic light of wavelength `lambda,` the intensity of light at a point on the screen where path difference is `lambda`, is K units. What is the intensity of lgight at a point where path difference is `lambda/3`.

Options

A. `(K)/(4)`
B. `(K)/(3)`
C. `(K)/(2)`
D. K

Answer

To solve the problem, we need to find the intensity of light at a point where the path difference is \( \frac{\lambda}{3} \) given that the intensity at a point where the path difference is \( \lambda \) is \( K \) units.

### Step-by-Step Solution:

1. **Understanding the relationship between path difference and phase difference**:
The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

2. **Calculate the phase difference for path difference \( \lambda \)**:
For a path difference of \( \lambda \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]
This corresponds to a complete cycle, and the intensity at this point is given as \( K \).

3. **Determine the intensity formula**:
The resultant intensity \( I \) when two waves of equal intensity \( I_0 \) interfere is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \]
Since both waves have the same intensity, we can denote \( I_1 = I_2 = I_0 \):
\[ I = 2I_0 + 2I_0 \cos(\Delta \phi) = 2I_0(1 + \cos(\Delta \phi)) \]

4. **Calculate the intensity for path difference \( \lambda \)**:
At \( \Delta \phi = 2\pi \):
\[ \cos(2\pi) = 1 \]
Therefore, the intensity becomes:
\[ I = 2I_0(1 + 1) = 4I_0 \]
Given \( I = K \), we have:
\[ K = 4I_0 \quad \Rightarrow \quad I_0 = \frac{K}{4} \]

5. **Calculate the phase difference for path difference \( \frac{\lambda}{3} \)**:
For a path difference of \( \frac{\lambda}{3} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \]

6. **Calculate the intensity for path difference \( \frac{\lambda}{3} \)**:
Substitute \( \Delta \phi = \frac{2\pi}{3} \) into the intensity formula:
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Thus, the intensity becomes:
\[ I = 2I_0\left(1 - \frac{1}{2}\right) = 2I_0 \cdot \frac{1}{2} = I_0 \]
Substituting \( I_0 = \frac{K}{4} \):
\[ I = \frac{K}{4} \]

### Final Answer:
The intensity of light at a point where the path difference is \( \frac{\lambda}{3} \) is \( \frac{K}{4} \).