Q92 / 120 · VITEEE 2018 · Physics

In Young's double slit experiment intensity at a point is $(1 / 4)$ of the maximum intensity. Angular position of this point is (separation between slits is $d$ )

  • A
    $\sin ^{-1}(\lambda / d)$
  • B
    $\sin ^{-1}(\lambda / 2 d)$
  • Correct answer
    C
    $\sin ^{-1}(\lambda / 3 d)$
  • D
    $\sin ^{-1}(\lambda / 4 d)$

Solution

If $a$ is the amplitude of the wave then $\frac{I_{\max }}{4}=a^{2}=a^{2}+a^{2}+2 a a \cos \phi$ or $\quad \cos \phi=-\frac{1}{2}$ or $\quad \phi=\frac{2 \pi}{3}$. Corresponding path difference, $\Delta x=\frac{\phi \times \lambda}{2 \pi}=\frac{(2 \pi / 3) \times \lambda}{2 \pi}=\frac{\lambda}{3}$ So $\quad d \sin \theta=\frac{\lambda}{3}$ or $\theta=\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$

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

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