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(B) Given: Intensities of two beams of monochromatic light are $I_1 = 64 \ mW$ and $I_2 = 4 \ mW$.The formula for the resultant intensity of two inter

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$60^\circ$

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(B) Given: Intensities of two beams of monochromatic light are $I_1 = 64 \ mW$ and $I_2 = 4 \ mW$.The formula for the resultant intensity of two interfering waves is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi$.When the intensity is reduced to $84 \ mW$,we substitute the known values into the equation:$84 = 64 + 4 + 2 \sqrt{64 	imes 4} \cos \phi$$84 = 68 + 2 	imes 8 	imes 2 \cos \phi$$84 - 68 = 32 \cos \phi$$16 = 32 \cos \phi$$\cos \phi = \frac{16}{32} = \frac{1}{2}$Therefore,$\phi = \cos^{-1} \left( \frac{1}{2} ight) = 60^\circ$.

Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

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