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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 $d$ )
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Verified Answer
The correct answer is:
$\sin ^{-1}(\lambda / 3 d)$
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
$\theta=\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
$\frac{I_{\max }}{4}=a^{2}=a^{2}+a^{2}+2 a a \cos \phi$
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
$\theta=\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
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