In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength…

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:
  1. 0.75
  2. 1.5
  3. 3.0
  4. 4.5

Solution

$\lambda=6000 Ã…=6 \times 10^{-7} \mathrm{~m}$ Path difference for dark fringe; $\Delta x=(2 n+1) \lambda / 2$ For third dark fringe, $n=2$ $\therefore \quad \Delta x=(2 \times 2+1) \times \frac{6 \times 10^{-7}}{2}$ $=\frac{5 \times 6 \times 10^{-7}}{2}$ $=15 \times 10^{-7}$ $=1.5 \times 10^{-6} \mathrm{~m}$ $=1.5 \mu \mathrm{m}$

Asked in: AP EAMCET 2003

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