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 $\phi=\frac{2 \pi}{\lambda} y \quad \ldots(i)$
We also know that for constructive interference to $\quad \phi=2 n \pi \ldots(ii)$
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 $\frac{2 \pi}{\lambda} y=(2 n+1) \pi$
$\therefore$ $y=(2 n+1) \frac{\lambda}{2}$
For $n$th minima : $y=y_n$
$\therefore$$y_n=(2 n+1) \frac{\lambda}{2}$
Thus, for destructive interference, path difference must be odd multiple of $\frac{\lambda}{2}$.
Phase difference $=\frac{2 \pi}{\lambda} \times$ path difference
or $\phi=\frac{2 \pi}{\lambda} y \quad \ldots(i)$
We also know that for constructive interference to $\quad \phi=2 n \pi \ldots(ii)$
From equation (i) and (ii)
$\frac{2 \pi}{\lambda} y=2 n \pi$
or $y=2 n \cdot \frac{\lambda}{2}$
For $n$th maxima, $y=y_n$
$\underline {y_n=2 n \cdot \frac{\lambda}{2}}$
Thus, to occur the constructive interference, path difference must be even multiple of $\frac{\lambda}{2}$
For minimum intensity $\quad \phi=(2 n+1) \pi$
or $\frac{2 \pi}{\lambda} y=(2 n+1) \pi$
$\therefore$ $y=(2 n+1) \frac{\lambda}{2}$
For $n$th minima : $y=y_n$
$\therefore$$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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