The fringe width in an interference pattern is ' $X$ '. The distance between the sixth dark fringe from one…

The fringe width in an interference pattern is ' $X$ '. The distance between the sixth dark fringe from one side of central bright band to the fourth bright fringe on other side is
  1. 1.5 X
  2. 2 X
  3. 5.5 X
  4. 9.5 X

Solution

Fringe width is given by, $W=\frac{\lambda D}{d}=X$ ...(given) Position of $4^{\text {th }}$ bright fringe $=n \frac{\lambda D}{d}=4 X$ $\begin{aligned} \text { Position of } 6^{\text {th }} \text { dark fringe } & =(2 \mathrm{n}-1) \frac{\lambda \mathrm{D}}{2 \mathrm{~d}} \\ & =(2(6)-1) \frac{\mathrm{X}}{2} \\ & =5.5 \mathrm{X} \end{aligned}$ $\therefore \quad$ Total distance $=(4+5.5) \mathrm{X}=9.5 \mathrm{X}$

Asked in: MHT CET 2024 (09 May Shift 2)

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