Q150 / 160 · TS EAMCET 2011 · Physics

In the Young's double slit experiment, the resultant intensity at a point on the screen is $75 \%$ of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • Correct answer
    C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$

Solution

$\begin{aligned} & \text { Given, } I_R=75 \% \text { of } I_{\max } \\ &=\frac{3}{4} I_{\max } \\ &=\frac{3}{4}\left(4 a^2\right)=3 a^2 \\ & \Rightarrow \quad 4 a^2 \cos ^2 \frac{\phi}{2}=3 a^2 \\ & \cos ^2 \frac{\phi}{2}=\frac{3}{4} \text { or } \cos \frac{\phi}{2}=\frac{\sqrt{3}}{2} \\ & \Rightarrow \quad \frac{\phi}{2}=\frac{\pi}{6} \text { or } \phi=\frac{\pi}{3}\end{aligned}$

Hence, the correct answer is option: 'C'.

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