Two light waves of intensities I_0 and 9I_0 s | vedclass

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(A) Given intensities are $I_1 = I_0$ and $I_2 = 9I_0$. The resultant intensity is $I_R = 7I_0$.The formula for resultant intensity in interference is

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

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(A) Given intensities are $I_1 = I_0$ and $I_2 = 9I_0$. The resultant intensity is $I_R = 7I_0$.The formula for resultant intensity in interference is given by:$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta\phi)$Substituting the given values:$7I_0 = I_0 + 9I_0 + 2\sqrt{I_0 \cdot 9I_0} \cos(\Delta\phi)$$7I_0 = 10I_0 + 2(3I_0) \cos(\Delta\phi)$$7I_0 - 10I_0 = 6I_0 \cos(\Delta\phi)$$-3I_0 = 6I_0 \cos(\Delta\phi)$$\cos(\Delta\phi) = -\frac{3I_0}{6I_0} = -\frac{1}{2}$Since $\cos(\Delta\phi) = -\frac{1}{2}$, the phase difference $\Delta\phi$ is $120^{\circ}$ or $\frac{2\pi}{3}$ radians.

Two light waves of intensities $I_0$ and $9I_0$ superpose at a point to produce a resultant intensity of $7I_0$. Calculate the phase difference between the light waves.

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