Question

In an interference experiment, the phase difference between the waves reaching a first dark point is

• A. zero
• B. \(\pi ^{\mathrm{c}}\)
• C. \((\frac{3\pi}{2})^{\mathrm{c}}\)
• D. \((2\pi )^{c}\)

Hint:

A dark fringe needs destructive interference with the smallest odd multiple of pi.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Destructive interference happens when the phase difference is an odd multiple of \(\pi\).

Step 2: First dark point
The first dark fringe corresponds to the smallest such value, \(\phi=(2n+1)\pi\) with \(n=0\).

Step 3: Result
So the phase difference is \(\pi\) radian.

Final Answer:
The phase difference at the first dark point is pi radian. \[ \boxed{\text{(B) }\pi^{c}} \]