In the Young's double slit experiment using a monochromatic light of wavelength λ , the path difference…

In the Young's double slit experiment using a monochromatic light of wavelength λ , the path difference ( in terms of an integer n ) corresponding to any point having half the peak intensity is
  1. $(2n+1)\frac{\lambda}{2}$
  2. $ (2n+1) \frac{\lambda}{4} $
  3. $(2n+1)\frac{\lambda}{8}$
  4. $(2n+1)\frac{\lambda}{16}$

Solution

$I_{\text{max}} = 4I$
$2I = 2I + 2I \cos \theta$
$\cos \theta = 0 \Rightarrow \theta = (2n+1) \frac{\pi}{2}$
$\frac{2\pi}{\lambda} x = (2n+1) \frac{\pi}{2}$
$x = \frac{(2n+1)\lambda}{4}$

Asked in: JEE Advanced 2013 (Paper 1)

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