Question

The ratio of the distance of $n^{th}$ bright band and $m^{th}$ dark band from the central bright band in an interference pattern is ______.

• A. $n : m$
• B. $m : n$
• C. $n : (m - 1/2)$
• D. $(n - 1/2) : m$

Hint:

The first dark band ($m=1$) happens at $0.5 \frac{\lambda D}{d}$. The formula $m - 0.5$ perfectly outputs $0.5$ when $m=1$. If you ever forget if it's $(m-0.5)$ or $(m+0.5)$, just check which one gives $0.5$ for the very first band!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), the distance of fringes from the central maxima depends on the path difference. Bright bands occur at integer multiples of fringe width, while dark bands occur at half-integer multiples.

Step 2: Formula Application:
Distance of $n^{th}$ bright band: $y_n = n \beta$, where $\beta = \frac{\lambda D}{d}$.
Distance of $m^{th}$ dark band: $y_m = (m - 1/2) \beta$.

Step 3: Explanation:
The ratio is $\frac{y_n}{y_m} = \frac{n \beta}{(m - 1/2) \beta} = \frac{n}{m - 1/2}$.

Step 4: Final Answer:
The ratio is $n : (m - 1/2)$.