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Question

(A) Let the phase difference introduced by the mica sheet be $\phi$. The effective path difference at a point on the screen is $\Delta x_{eff} = d \si

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Let the phase difference introduced by the mica sheet be $\phi$. The effective path difference at a point on the screen is $\Delta x_{eff} = d \sin 	heta - \frac{\phi \lambda}{2 \pi}$.For the second-order minima above the center, the path difference is $\Delta x = -\frac{3\lambda}{2}$. Thus, $d \sin 	heta_1 - \frac{\phi \lambda}{2 \pi} = -\frac{3\lambda}{2}$.For the second-order maxima below the center, the path difference is $\Delta x = 2\lambda$. Thus, $d \sin 	heta_2 - \frac{\phi \lambda}{2 \pi} = 2\lambda$.Since the mica sheet shifts the pattern upward, the phase difference $\phi$ must satisfy the condition that the second-order minima (originally at $2.5\lambda$) and second-order maxima (originally at $2\lambda$) are shifted such that they appear on opposite sides of the center.This implies the phase difference $\phi$ must be in the range $3\pi Checking the options: $\frac{\pi}{3} \approx 1.04\pi$, $\frac{7\pi}{2} = 3.5\pi$, $\frac{10\pi}{3} \approx 3.33\pi$, $\frac{11\pi}{3} \approx 3.66\pi$.Only $\frac{\pi}{3}$ lies outside the range $(3\pi, 4\pi)$.

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