Question

In Young's double slit experiment, for the \(n^{\text{th}}\) dark fringe (\(n = 1,2,3,\ldots\)) the phase difference of the interfering waves in radian will be

• A. \(n\cdot \frac{π}{2}\)
• B. \((2n+1)π\)
• C. \((2n-1)π\)
• D. \((2n-1)\frac{π}{2}\)

Hint:

Dark fringes occur at path difference (2n - 1) lambda/2, and phase difference is 2 pi/lambda times path difference.

Answer 1

The Correct Option is C

Step 1: Path Difference:
For the \(n\)th dark fringe (\(n=1,2,3,\ldots\)), the path difference is \(\Delta=(2n-1)\dfrac\lambda2\).

Step 2: Phase Difference:
\[ \phi=\frac{2\pi}\lambda\Delta=\frac{2\pi}\lambda(2n-1)\frac\lambda2=(2n-1)\pi \]

Step 3: Check the Options:
For \(n=1\) the first dark fringe has phase \(\pi\). Option (C) gives \(\pi\). Option (B) \((2n+1)\pi\) gives \(3\pi\) for \(n=1\), which is the second dark fringe. Option (D) \((2n-1)\pi/2\) gives \(\pi/2\) for \(n=1\), which is not destructive. Option (A) \(n\pi/2\) gives \(\pi/2\) too.

Final Answer:
The phase difference is \((2n-1)\pi\), option (C). \[ \boxed{\text{(C) } (2n-1)\pi} \]