At two point P and Q on screen in Young's double slit experiment, waves from slits `S_(1)` and `S_(2)` have a path difference of 0 and `(lamda)/(4)` respectively. The ratio of intensities at P and Q will be:

Options

A. `3:2`
B. `2:1`
C. `sqrt(2):1`
D. `4:1`

Explanation

To solve the problem, we need to determine the ratio of intensities at points P and Q in a Young's double slit experiment, given the path differences from the slits S1 and S2.

### Step-by-Step Solution:

1. **Identify the Path Differences**:
- At point P, the path difference is \(0\).
- At point Q, the path difference is \(\frac{\lambda}{4}\).

2. **Calculate the Phase Difference at Point P**:
- The phase difference \(\Delta \phi_P\) at point P can be calculated using the formula:
\[ \Delta \phi_P = \frac{2\pi}{\lambda} \times \text{(path difference)} \]
- Since the path difference is \(0\):
\[ \Delta \phi_P = \frac{2\pi}{\lambda} \times 0 = 0 \]

3. **Calculate the Resultant Intensity at Point P**:
- The resultant intensity \(I_P\) at point P can be calculated using the formula:
\[ I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi_P) \]
- Assuming both slits have the same intensity \(I_1 = I_2 = I_0\):
\[ I_P = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(0) = 2I_0 + 2I_0 = 4I_0 \]

4. **Calculate the Phase Difference at Point Q**:
- The phase difference \(\Delta \phi_Q\) at point Q is:
\[ \Delta \phi_Q = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]

5. **Calculate the Resultant Intensity at Point Q**:
- The resultant intensity \(I_Q\) at point Q can be calculated as:
\[ I_Q = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi_Q) \]
- Substituting the values:
\[ I_Q = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{\pi}{2}\right) = 2I_0 + 0 = 2I_0 \]

6. **Calculate the Ratio of Intensities**:
- The ratio of intensities at points P and Q is:
\[ \frac{I_P}{I_Q} = \frac{4I_0}{2I_0} = 2 \]
- Therefore, the ratio of intensities can be expressed as:
\[ \text{Ratio} = 2:1 \]

### Final Answer:
The ratio of intensities at points P and Q is \(2:1\).

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