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. `1:1`
B. `1:2`
C. `2:1`
D. `1:4`

Answer

To solve the problem of finding the ratio of intensity at a point on the screen where the path difference between two interfering waves is \( \frac{\lambda}{4} \) and that at the central fringe, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Intensity Formula**:
For two interfering waves of equal intensity \( I_0 \), the intensity \( I \) at a point on the screen can be given by:
\[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the two waves.

2. **Calculate the Phase Difference**:
The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
where \( \Delta x \) is the path difference. Given that \( \Delta x = \frac{\lambda}{4} \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

3. **Substitute Phase Difference into Intensity Formula**:
Now substitute \( \phi = \frac{\pi}{2} \) into the intensity formula:
\[ I_1 = 4I_0 \cos^2\left(\frac{\pi/2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{4}\right) \]
We know that \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_1 = 4I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \cdot \frac{1}{2} = 2I_0 \]

4. **Calculate Intensity at the Central Fringe**:
At the central fringe, the phase difference \( \phi = 0 \):
\[ I_2 = 4I_0 \cos^2\left(0\right) = 4I_0 \cdot 1^2 = 4I_0 \]

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

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

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