In a YDSE with two identical slits,when the u | vedclass

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(A) Let the phase difference introduced by the mica sheet be $\phi$. The intensity at the center of the screen is given by $I = I_0 \cos^2(\phi/2)$. G

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(A) Let the phase difference introduced by the mica sheet be $\phi$. The intensity at the center of the screen is given by $I = I_0 \cos^2(\phi/2)$. Given $I = 0.75 I_0$,we have $\cos^2(\phi/2) = 3/4$,which implies $\cos(\phi/2) = \pm \sqrt{3}/2$. Thus,$\phi/2 = n\pi \pm \pi/6$,or $\phi = 2n\pi \pm \pi/3$. For $n=1$,$\phi = 2\pi \pm \pi/3$,giving $\phi = 5\pi/3$ or $7\pi/3$. For $n=2$,$\phi = 4\pi \pm \pi/3$,giving $\phi = 11\pi/3$ or $13\pi/3$. For $n=3$,$\phi = 6\pi \pm \pi/3$,giving $\phi = 17\pi/3$ or $19\pi/3$. The condition that the second minimum is above the center and the third maximum is below the center constrains the value of $\phi$. Based on the shift analysis,the phase difference $\phi$ must satisfy $3\pi < \phi < 4\pi$. Among the options,$\pi/3$ does not fall in the required range.

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