Question

In Young's double slit experiment, the distance of $n^{\text{th}}$ dark band from the central bright band in terms of bandwidth $\beta$ is

• A. $n\beta$
• B. $(n - 1)\beta$
• C. $(n - 0.5)\beta$
• D. $(n + 0.5)\beta$

Hint:

To verify fringe position formulas instantly, plug in $n=1$ for the first dark band. The first dark fringe lies exactly halfway between the central maximum ($0$) and the first bright fringe ($\beta$), which means its position must be $0.5\beta$. Substituting $n=1$ into $(n-0.5)\beta$ perfectly yields $0.5\beta$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the mathematical formula representing the position of the $n^{\text{th}}$ dark fringe (destructive interference minimum) relative to the central bright maximum in a standard Young's Double Slit Experiment (YDSE).

Step 2: Key Formula or Approach:
In YDSE, destructive interference happens at positions where the path difference is an odd multiple of half-wavelengths:
$$\Delta x = (2n - 1)\frac{\lambda}{2}$$ where $n = 1, 2, 3, \dots$ represents the order of the dark fringe.
The corresponding position on the screen is given by $y_n = \Delta x \frac{D}{d}$, and the fringe width (bandwidth) is defined as:
$$\beta = \frac{\lambda D}{d}$$

Step 3: Detailed Explanation:
Let's express the position equation of the dark band in terms of the fringe width $\beta$:
$$y_n = (2n - 1)\frac{\lambda D}{2d} = \left(\frac{2n - 1}{2}\right)\frac{\lambda D}{d}$$ Distributing the denominator of $2$ inside the parentheses:
$$y_n = \left(\frac{2n}{2} - \frac{1}{2}\right)\beta = (n - 0.5)\beta$$ This shows that the $n^{\text{th}}$ dark band is located at a distance of $(n - 0.5)\beta$ from the center.

Step 4: Final Answer:
The distance is $(n - 0.5)\beta$, which corresponds to option (C).