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: Identify the quantity.
We want the distance of the $n^{\text{th}}$ dark band from the central bright fringe, written using the band width $\beta$.
Step 2: Condition for a dark band.
Darkness (destructive interference) occurs when the path difference is an odd multiple of half a wavelength: $\Delta x = (2n - 1)\dfrac{\lambda}{2}$, with $n = 1, 2, 3, \dots$
Step 3: Position from path difference.
On the screen, position $y_n = \Delta x \cdot \dfrac{D}{d}$, where $D$ is the slit-screen distance and $d$ the slit separation.
Step 4: Bring in the band width.
The fringe width is $\beta = \dfrac{\lambda D}{d}$. So $y_n = (2n - 1)\dfrac{\lambda}{2}\cdot\dfrac{D}{d} = \dfrac{(2n-1)}{2}\,\beta$.
Step 5: Simplify the coefficient.
$\dfrac{2n - 1}{2} = n - \dfrac{1}{2} = (n - 0.5)$. Hence $y_n = (n - 0.5)\beta$.
Step 6: Check with the first band.
For $n = 1$, $y_1 = 0.5\beta$, exactly halfway to the first bright fringe, which is correct. So the answer is option (3).
\[ \boxed{y_n = (n - 0.5)\beta} \]