Question

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

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

Hint:

Bright fringes = Even multiples of $\pi$ ($2n\pi$). Dark fringes = Odd multiples of $\pi$ ($(2n-1)\pi$).

Answer 1

The Correct Option is C

Step 1: Concept
Dark fringes (destructive interference) occur when the path difference is an odd multiple of $\lambda/2$ and the phase difference is an odd multiple of $\pi$.

Step 2: Meaning
Phase difference ($\phi$) and path difference ($\Delta x$) are related by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$.

Step 3: Analysis
For the $n$th dark fringe, the path difference is $\Delta x = (2n-1)\frac{\lambda}{2}$. Substituting this into the phase formula: $\phi = \frac{2\pi}{\lambda} \times (2n-1)\frac{\lambda}{2} = (2n-1)\pi$.

Step 4: Conclusion
The phase difference for the $n$th dark fringe is $(2n-1)\pi$. Final Answer: (C)