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In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength $6000 Ã…$ coming from the coherent sources $S_1$ and $S_2$. At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_1 P-S_2 P$ in microns is:
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The correct answer is:
1.5
$\lambda=6000 Ã…=6 \times 10^{-7} \mathrm{~m}$
Path difference for dark fringe;
$\Delta x=(2 n+1) \lambda / 2$
For third dark fringe, $n=2$
$\therefore \quad \Delta x=(2 \times 2+1) \times \frac{6 \times 10^{-7}}{2}$
$=\frac{5 \times 6 \times 10^{-7}}{2}$
$=15 \times 10^{-7}$
$=1.5 \times 10^{-6} \mathrm{~m}$
$=1.5 \mu \mathrm{m}$
Path difference for dark fringe;
$\Delta x=(2 n+1) \lambda / 2$
For third dark fringe, $n=2$
$\therefore \quad \Delta x=(2 \times 2+1) \times \frac{6 \times 10^{-7}}{2}$
$=\frac{5 \times 6 \times 10^{-7}}{2}$
$=15 \times 10^{-7}$
$=1.5 \times 10^{-6} \mathrm{~m}$
$=1.5 \mu \mathrm{m}$
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