The wavelength of the light used in Young's double slit experiment is $\lambda$. The intensity at a point on the screen is $I$,where the path difference is $\lambda/6$. If $I_0$ denotes the maximum intensity,then the ratio of $I$ and $I_0$ is

  • A
    $0.866$
  • B
    $0.5$
  • C
    $0.707$
  • D
    $0.75$

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