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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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Maximum intensity of the interference pattern in a young double slit experiment is I. Distance between slit is $d= 5 \lambda$ where $\lambda $ is wave length of the light source. What will be the intensity of light at a point directly in front of one of the slits. $(D= 50 \lambda)$

$(a)\;\frac{I}{2} \\ (b)\;\frac{I}{3} \\ (c)\;\frac{3I}{2} \\ (d)\;\frac{I}{4} $

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Path difference $\Delta =\large\frac{yd}{D}$
$d= 5 \lambda$
$D= 50 \lambda$
$y= \large\frac{d}{2}$
$\quad= \large\frac{5}{2}$$ \lambda$
$\therefore \Delta x= \large\frac{\lambda}{4}$
Corresponding phase difference $\phi$
$\phi =\large\frac{2 \pi}{\lambda} $$\Delta x$
$\qquad= \large\frac{2 \pi }{\lambda} \times \frac{\lambda}{4}$
$\large\frac{\phi}{2} =\frac{\pi}{4}$
$I_1= I \cos^2 \large\frac{\phi}{2}$
$\qquad= I \cos^2 \large\frac{\pi}{4}$
$\qquad= I ( \large\frac{ 1}{\sqrt 2})^2$
$\qquad =\large\frac{ I}{2}$
Hence a is the correct answer.
answered Jan 23, 2014 by meena.p

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