Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics1 Mark
Question
Write the upper and lower limits of interference in terms of path difference.
✓
Answer
We know that : Phase difference $=\frac{2 \pi}{\lambda} \times$ path difference or $\phi=\frac{2 \pi}{\lambda} y \ldots(i) $ We also know that for constructive interference to $\phi=2 n \pi \ldots(ii) $ From equation (i) and (ii) $ \frac{2 \pi}{\lambda} y=2 n \pi $ or $ y=2 n \cdot \frac{\lambda}{2} $ For $n$th maxima, $y=y_n$ $\underline {y_n=2 n \cdot \frac{\lambda}{2}}$ Thus, to occur the constructive interference, path difference must be even multiple of $\frac{\lambda}{2}$ For minimum intensity $\quad \phi=(2 n+1) \pi$ or $\quad\frac{2 \pi}{\lambda} y=(2 n+1) \pi$ $\therefore$$\quad y=(2 n+1) \frac{\lambda}{2}$ For $n$th minima : $y=y_n$ $\therefore$ $y_n=(2 n+1) \frac{\lambda}{2}$ Thus, for destructive interference, path difference must be odd multiple of $\frac{\lambda}{2}$.
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