In a Young's double slit experiment , the intensity of light at a point on the screen where the path difference is `lambda` is k units. Find the intensity at a point where the path difference is (a) `(lambda)/(4)` (b) `(lambda)/(3)` and (c) `(lambda)/(2)`

Answer

To solve the problem, we need to find the intensity of light at different path differences in a Young's double slit experiment. We know that the intensity \( I \) at a point on the screen can be expressed in terms of the phase difference \( \phi \) as follows:

\[ I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]

where \( I_0 \) is the maximum intensity.

### Given:
- Intensity at path difference \( \lambda \) is \( k \) units.
- We need to find the intensity at path differences \( \frac{\lambda}{4} \), \( \frac{\lambda}{3} \), and \( \frac{\lambda}{2} \).

### Step-by-Step Solution:

1. **Determine the Phase Difference for Path Difference \( \lambda \)**:
- The phase difference \( \phi \) is given by:
\[ \phi = \frac{2\pi}{\lambda} \times \text{(path difference)} \]
- For a path difference of \( \lambda \):
\[ \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi \]
- At this point, we know that:
\[ I = 4 I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4 I_0 \cos^2(\pi) = 4 I_0 \cdot 1 = 4 I_0 \]
- Given that this intensity is \( k \):
\[ 4 I_0 = k \implies I_0 = \frac{k}{4} \]

2. **(a) Path Difference \( \frac{\lambda}{4} \)**:
- Calculate the phase difference:
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]
- Substitute into the intensity formula:
\[ I = 4 I_0 \cos^2\left(\frac{\pi}{4}\right) = 4 I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = 4 I_0 \cdot \frac{1}{2} = 2 I_0 \]
- Substitute \( I_0 \):
\[ I = 2 \cdot \frac{k}{4} = \frac{k}{2} \]

3. **(b) Path Difference \( \frac{\lambda}{3} \)**:
- Calculate the phase difference:
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \]
- Substitute into the intensity formula:
\[ I = 4 I_0 \cos^2\left(\frac{2\pi}{6}\right) = 4 I_0 \cos^2\left(\frac{\pi}{3}\right) = 4 I_0 \cdot \left(\frac{1}{2}\right)^2 = 4 I_0 \cdot \frac{1}{4} = I_0 \]
- Substitute \( I_0 \):
\[ I = \frac{k}{4} \]

4. **(c) Path Difference \( \frac{\lambda}{2} \)**:
- Calculate the phase difference:
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \]
- Substitute into the intensity formula:
\[ I = 4 I_0 \cos^2\left(\frac{\pi}{2}\right) = 4 I_0 \cdot 0 = 0 \]

### Summary of Results:
- (a) Intensity at path difference \( \frac{\lambda}{4} \) is \( \frac{k}{2} \).
- (b) Intensity at path difference \( \frac{\lambda}{3} \) is \( \frac{k}{4} \).
- (c) Intensity at path difference \( \frac{\lambda}{2} \) is \( 0 \).