KCET · Physics · Wave Optics
For the constructive interference the path difference between the two interfering waves must be equal to
- A \((2 \mathrm{n}+1) \lambda\)
- B \(2 \mathrm{n} \pi\)
- C \(\mathrm{n} \lambda\)
- D \((2 \mathrm{n}+1) \frac{\lambda}{2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{n} \lambda\)
Step-by-step Solution
Detailed explanation
Phase difference,
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\)
In a constructive interference,
\(\begin{array}{cc}
\Delta \phi=2 \mathrm{n} \pi & \text { where } \mathrm{n}=0,1,2,3, \ldots \\
\therefore \quad 2 \mathrm{n} \pi=\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\
\text { or } \quad & \Delta \mathrm{x}=\mathrm{n} \lambda
\end{array}\)
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\)
In a constructive interference,
\(\begin{array}{cc}
\Delta \phi=2 \mathrm{n} \pi & \text { where } \mathrm{n}=0,1,2,3, \ldots \\
\therefore \quad 2 \mathrm{n} \pi=\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\
\text { or } \quad & \Delta \mathrm{x}=\mathrm{n} \lambda
\end{array}\)
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