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In YDSE intensity at a point is $\bigg (\large\frac{1}{4} \bigg) $ of maximum intensity. Angular position of this point is :

$(a)\;\sin ^{-1} \bigg( \frac{\lambda}{d}\bigg) \\ (b)\;\sin ^{-1} \bigg( \frac{\lambda}{3d}\bigg) \\ (c)\;\sin ^{-1} \bigg( \frac{\lambda}{2d}\bigg) \\ (d)\;\sin ^{-1} \bigg( \frac{\lambda}{4d}\bigg)$

1 Answer

$I= I_{max} \cos ^2 \bigg(\large\frac{\phi}{2}\bigg)$
$\therefore \large\frac{I_{max}}{4}$$ =I_{max} \cos ^2 \large\frac{\phi}{2}$
$\cos \large\frac{\phi}{2} =\frac{1}{2}$
or $ \large \frac {\phi}{2}= \frac{\pi}{3}$
$\therefore \phi = \large\frac{2 \pi}{3}$
$\qquad = \bigg( \large\frac{2 \pi}{\lambda}\bigg)$$.\Delta x$
Where $\Delta x =d \sin \theta$
Substituting in equation we get,
$\sin \theta= \large\frac{\lambda}{3d}$
or $\theta= \sin ^{-1} \bigg( \large\frac{\lambda}{3 d}\bigg)$
Hence b is the correct answer.
answered Feb 20, 2014 by meena.p

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