MCQ
Consider an $YDSE$ that has different slits width, as a result, amplitude of waves from two slits are $A$ and $2A$ , respectively. If $I_0$ be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is $\phi $ is
  • A
    ${I_0}{\cos ^2}\phi $
  • B
    $\frac{{{I_0}}}{3}{\sin ^2}\frac{\phi }{2}$
  • $\frac{{{I_0}}}{9}\left[ {5 + 4\,\cos \phi } \right]$
  • D
    $\frac{{{I_0}}}{9}\left[ {5 + 8\,\cos \phi } \right]$

Answer

Correct option: C.
$\frac{{{I_0}}}{9}\left[ {5 + 4\,\cos \phi } \right]$
c
As amplitude are $A$ and $2 A,$ so intensities would be in the ratio $1: 4,$ let us say $I$ and $4I$

$I_{\max }=I_{0}=I+4 I+2 \sqrt{4 I^{2}}=9 I$

$\Rightarrow I=\frac{I_{0}}{9}$

Intensity at any point,

$I^{\prime}=I+4 I+2 \sqrt{4 I^{2}} \cos \phi$

$\Rightarrow I^{\prime}=5 I+4 I \cos \phi=\frac{I_{0}}{9}(5+4 \cos \phi)$

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