Step 1: Understanding the Concept:
In Young's double-slit experiment, an interference pattern consisting of alternate bright and dark fringes is formed on the screen.
Destructive interference occurs when the waves from the two slits arrive out of phase, producing a dark fringe.
Step 2: Key Formula or Approach:
For destructive interference (dark fringes), the phase difference $\Delta \phi$ between the two interfering waves must be an odd multiple of $\pi$.
Mathematically, this is expressed as:
\[ \Delta \phi = (2n - 1)\pi \quad \text{where } n = 1, 2, 3, \dots \]
Alternatively, it can be written as $(2n + 1)\pi$ if $n = 0, 1, 2, \dots$.
Step 3: Detailed Explanation:
The problem specifies the index sequence as $n = 1, 2, 3 \dots$.
- For the 1st dark fringe ($n=1$), the phase difference should be $\pi$. Substituting $n=1$ into $(2n-1)\pi$ gives $(2(1)-1)\pi = \pi$.
- For the 2nd dark fringe ($n=2$), the phase difference should be $3\pi$. Substituting $n=2$ into $(2n-1)\pi$ gives $(2(2)-1)\pi = 3\pi$.
This perfectly matches the standard sequence for odd multiples of $\pi$ starting from $\pi$.
Let's check option (B) which is $(2n+1)\pi$: If $n=1$, it gives $3\pi$, which is the 2nd dark fringe, not the 1st. So it is incorrect for the given index $n=1,2,3\dots$.
Step 4: Final Answer:
The phase difference for the $n$th dark fringe is $(2n-1)\pi$.