In Young's double slit experiment, the intensity of light at a point on the screen where path difference is `lambda` is I. If intensity at another point is I/4, then possible path differences at this point are

Options

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

Explanation

To solve the problem, we need to find the possible path differences at a point on the screen in Young's double slit experiment where the intensity of light is given as I/4, while at a path difference of λ, the intensity is I.

### Step-by-Step Solution:

1. **Understanding Intensity in Young's Experiment**:
The intensity (I) at any point on the screen in Young's double slit experiment is related to the phase difference (φ) between the two waves arriving at that point. The relationship is given by:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_{\text{max}} \) is the maximum intensity.

2. **Finding Phase Difference for Given Intensity**:
At a point where the path difference is λ, the phase difference is:
\[ \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]
At this phase difference, the intensity is:
\[ I = I_{\text{max}} \cos^2\left(\frac{2\pi}{2}\right) = I_{\text{max}} \cos^2(\pi) = I_{\text{max}} \cdot 1 = I \]

3. **Setting Up the Equation for I/4**:
Now, we want to find the path differences where the intensity is \( \frac{I}{4} \):
\[ \frac{I}{4} = I_{\text{max}} \cos^2\left(\frac{\phi'}{2}\right) \]
This implies:
\[ \cos^2\left(\frac{\phi'}{2}\right) = \frac{1}{4} \]

4. **Solving for Phase Difference**:
Taking the square root gives:
\[ \cos\left(\frac{\phi'}{2}\right) = \frac{1}{2} \]
The angles for which cosine equals \( \frac{1}{2} \) are:
\[ \frac{\phi'}{2} = \frac{\pi}{3} \quad \text{or} \quad \frac{\phi'}{2} = \frac{5\pi}{3} \]
Therefore:
\[ \phi' = \frac{2\pi}{3} \quad \text{or} \quad \phi' = \frac{10\pi}{3} \]

5. **Finding Path Differences**:
Now we can relate the phase difference back to the path difference using:
\[ \phi' = \frac{2\pi}{\lambda} \Delta x \]
Rearranging gives:
\[ \Delta x = \frac{\phi' \lambda}{2\pi} \]
Substituting the values of \( \phi' \):
- For \( \phi' = \frac{2\pi}{3} \):
\[ \Delta x = \frac{\frac{2\pi}{3} \lambda}{2\pi} = \frac{\lambda}{3} \]
- For \( \phi' = \frac{10\pi}{3} \):
\[ \Delta x = \frac{\frac{10\pi}{3} \lambda}{2\pi} = \frac{5\lambda}{3} \]

6. **Final Result**:
The possible path differences where the intensity is \( \frac{I}{4} \) are:
\[ \Delta x = \frac{\lambda}{3} \quad \text{and} \quad \Delta x = \frac{5\lambda}{3} \]