A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle `theta=sin^-1(lamda/(2d))` with the normal to the plane of the slits, there will be a dark fringe at the centre Po of the pattern.

Explanation

To solve the problem, we need to analyze the Young's double slit experiment with the given conditions. Here’s a step-by-step solution:

### Step 1: Understanding the Setup
In a Young's double slit experiment, we have two slits (S1 and S2) separated by a distance \( d \). A monochromatic light beam is incident on these slits at an angle \( \theta \) with respect to the normal (perpendicular) to the plane of the slits.

### Step 2: Determine the Path Difference
When light passes through the two slits, it travels different distances to reach a point on the screen. The path difference \( \Delta x \) between the light coming from S1 and S2 can be expressed in terms of the angle \( \theta \):
\[ \Delta x = d \sin \theta \]

### Step 3: Substitute the Given Angle
We are given that the angle \( \theta \) is defined as:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{2d}\right) \]
Substituting this into the equation for path difference, we have:
\[ \sin \theta = \frac{\lambda}{2d} \]
Thus, substituting this into the path difference equation gives:
\[ \Delta x = d \cdot \frac{\lambda}{2d} = \frac{\lambda}{2} \]

### Step 4: Condition for Dark Fringe
In order to have a dark fringe (destructive interference) at the center of the pattern, the path difference must be an odd multiple of \( \frac{\lambda}{2} \):
\[ \Delta x = \left(n + \frac{1}{2}\right) \lambda \]
For the first dark fringe, we can set \( n = 0 \):
\[ \Delta x = \frac{1}{2} \lambda \]

### Step 5: Conclusion
Since we have calculated that the path difference \( \Delta x = \frac{\lambda}{2} \), it satisfies the condition for a dark fringe at the center \( P_0 \) of the pattern. Therefore, we conclude that there will indeed be a dark fringe at the center of the interference pattern.

### Final Statement
Thus, we have shown that if the incident beam makes an angle \( \theta = \sin^{-1}\left(\frac{\lambda}{2d}\right) \) with the normal to the plane of the slits, there will be a dark fringe at the center \( P_0 \) of the pattern.

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