MHT CET · Physics · Wave Optics
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
- A 1.5 X
- B 2 X
- C 5.5 X
- D 9.5 X
Answer & Solution
Correct Answer
(D) 9.5 X
Step-by-step Solution
Detailed explanation
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 \(=(4+5.5) \mathrm{X}=9.5 \mathrm{X}\)
...(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 \(=(4+5.5) \mathrm{X}=9.5 \mathrm{X}\)
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