Q115 / 150 · MHT CET 2025 (26 Apr Shift 1) · Physics

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

  • A
    n: m
  • B
    m: n
  • Correct answer
    C
    $n:\left(m-\frac{1}{2}\right)$
  • D
    $\left(n-\frac{1}{2}\right): m$

Solution

The distance from the central bright band to the nth bright fringe is $x_n = n\frac{\lambda D}{d}$, while the distance to the mth dark fringe is $x_m = \left(m - \frac{1}{2}\right)\frac{\lambda D}{d}$.

The ratio between these distances simplifies to $\frac{x_n}{x_m} = \frac{n}{m - \frac{1}{2}}$, yielding $n : \left(m - \frac{1}{2}\right)$.

This corresponds to option C.

Hence, the correct answer is option: 'C'.

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