The path difference between two interfering waves of equal intensities at a point on the screen is `lambda//4`. The ratio of intensity at this point and that at the central fringe will be

Options

A. (a) `1:1`
B. (b) `1:2`
C. (c) `2:1`
D. (d) `1:4`

Explanation

To solve the problem, we need to find the ratio of intensity at a point where the path difference between two interfering waves is \( \frac{\lambda}{4} \) and the intensity at the central fringe (where the path difference is zero).

### Step-by-Step Solution:

1. **Understanding the Intensity Formula**:
The intensity \( I \) at a point in an interference pattern created by two waves of equal intensity \( I_0 \) can be expressed as:
\[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the two waves.

2. **Calculating the Phase Difference**:
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Given that the path difference \( \Delta x = \frac{\lambda}{4} \), we can substitute this into the equation:
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

3. **Finding the Intensity at the Given Point**:
Now substituting \( \phi = \frac{\pi}{2} \) into the intensity formula:
\[ I_1 = 4I_0 \cos^2\left(\frac{\pi}{4}\right) \]
Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we have:
\[ I_1 = 4I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \cdot \frac{1}{2} = 2I_0 \]

4. **Finding the Intensity at the Central Fringe**:
At the central fringe, the path difference is zero, which means \( \phi = 0 \):
\[ I_2 = 4I_0 \cos^2\left(0\right) = 4I_0 \cdot 1 = 4I_0 \]

5. **Calculating the Ratio of Intensities**:
Now, we can find the ratio of the intensity at the point with path difference \( \frac{\lambda}{4} \) to the intensity at the central fringe:
\[ \text{Ratio} = \frac{I_1}{I_2} = \frac{2I_0}{4I_0} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer:
The ratio of intensity at the point where the path difference is \( \frac{\lambda}{4} \) to that at the central fringe is \( \frac{1}{2} \).