In a standard Young's double-slit experiment set-up, two point P and Q are marked on the screen Path difference corresponding to point P is `lambda//2` and for point Q it is `lambda//4`. Here `lambda` is wavelength of light being used. If `I_P and I_Q` are the resultant intensities at points P and Q, then `I_Q//I_P` is

Options

A. 2
B. `1//2`
C. `0`
D. infinity

Explanation

To solve the problem, we need to find the ratio of the resultant intensities at points P and Q in a Young's double-slit experiment, given the path differences at these points.

### Step-by-step Solution:

1. **Identify the Path Differences:**
- For point P, the path difference (Δx_P) is given as \( \frac{\lambda}{2} \).
- For point Q, the path difference (Δx_Q) is given as \( \frac{\lambda}{4} \).

2. **Calculate the Phase Difference:**
- The phase difference (φ) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
- For point P:
\[ \phi_P = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi \]
- For point Q:
\[ \phi_Q = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

3. **Calculate the Resultant Intensity at Point P (I_P):**
- The resultant intensity at any point in the interference pattern can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Assuming the intensities from each slit (I_1 and I_2) are equal, let \( I_1 = I_2 = I \).
- For point P:
\[ I_P = I + I + 2\sqrt{I \cdot I} \cos(\pi) = 2I + 2I(-1) = 2I - 2I = 0 \]

4. **Calculate the Resultant Intensity at Point Q (I_Q):**
- For point Q:
\[ I_Q = I + I + 2\sqrt{I \cdot I} \cos\left(\frac{\pi}{2}\right) = 2I + 2I(0) = 2I \]

5. **Calculate the Ratio of Intensities (I_Q/I_P):**
- Now, we can find the ratio:
\[ \frac{I_Q}{I_P} = \frac{2I}{0} \]
- Since \( I_P = 0 \), this ratio is undefined, which can be interpreted as approaching infinity.

### Conclusion:
The ratio \( \frac{I_Q}{I_P} \) is infinite.