Question
In a Young double slit experiment, the wavelength of incident of incident light is $6000\text{Å}$, the separation between slits $S_{1}$ and 5 cm the distance between slits plane and screen is 50 cm , as shown in the figure below. If the resultant intensity at P is equal to the intensity due to individual slits, the path different between interfering waves is
$\_\_\_\_$Å .
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Solution
Sol. [Key Concept: The resultant intensity equals a single slit's when $I = 4I_{0}\cos^{2}\frac{\Delta\varphi}{2} = I_{0}$, so $\Delta\varphi = \frac{2\pi}{3}$ and the path difference $= \frac{\lambda}{3}$.]
$${I_{P} = I_{0} = 4I_{0}\cos^{2}\frac{\Delta\phi}{2} }{cos\frac{\Delta\phi}{2} = \frac{1}{2} }{\frac{\Delta\phi}{2} = \frac{\pi}{3} }{\frac{2\pi}{\lambda} = \Delta x = \Delta\phi = \frac{2\pi}{3} }{\Delta x = \frac{\lambda}{3} = 2000{\text{ }A}^{\circ}}$$
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