In the above problem, minimum value of t for which the intensity at point P on the screen exactly in front of the upper slit becomes maximum.

Options

A. 0.167s
B. 0.5 s
C. 0.012 s
D. 0.3 s

Answer

To solve the problem regarding the minimum value of time \( t \) for which the intensity at point \( P \) on the screen exactly in front of the upper slit becomes maximum, we need to analyze the conditions for constructive interference in a double-slit experiment.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In a double-slit experiment, light waves from two slits interfere with each other. The intensity at a point on the screen depends on the path difference between the two waves reaching that point.

2. **Condition for Maximum Intensity**:
- For maximum intensity (constructive interference) at point \( P \), the path difference \( \Delta x \) between the two waves must be an integer multiple of the wavelength \( \lambda \):
\[ \Delta x = n\lambda \quad (n = 0, 1, 2, \ldots) \]

3. **Path Difference Calculation**:
- The path difference can be expressed in terms of the distance traveled by the light from the slits to point \( P \). If the distance from the slits to point \( P \) is \( d \) and the angle of incidence is \( \theta \), then:
\[ \Delta x = d \sin \theta \]

4. **Time Calculation**:
- The time \( t \) taken for light to travel a distance \( d \) is given by:
\[ t = \frac{d}{v} \]
where \( v \) is the speed of light.

5. **Finding Minimum Time**:
- To find the minimum time \( t \) for which the intensity at point \( P \) becomes maximum, we need to ensure that the path difference corresponds to the first order maximum (i.e., \( n = 1 \)):
\[ d \sin \theta = \lambda \]
- Rearranging gives:
\[ d = \frac{\lambda}{\sin \theta} \]
- Substituting this into the time equation:
\[ t = \frac{\frac{\lambda}{\sin \theta}}{v} = \frac{\lambda}{v \sin \theta} \]

6. **Conclusion**:
- Therefore, the minimum value of time \( t \) for which the intensity at point \( P \) becomes maximum is:
\[ t = \frac{\lambda}{v \sin \theta} \]