Two light waves of intensities I and 4I super | vedclass

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(C) Let the amplitudes of the two waves be $A_1$ and $A_2$. Since intensity $I \propto A^2$,we have $I_1 = A_1^2 = I$ and $I_2 = A_2^2 = 4I$.Thus,$A_1

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$\sqrt{5}A$

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(C) Let the amplitudes of the two waves be $A_1$ and $A_2$. Since intensity $I \propto A^2$,we have $I_1 = A_1^2 = I$ and $I_2 = A_2^2 = 4I$.Thus,$A_1 = \sqrt{I}$ and $A_2 = \sqrt{4I} = 2\sqrt{I}$.The resultant amplitude $A_R$ of two waves with phase difference $\phi$ is given by $A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.Given $\phi = \pi / 2$,we have $\cos(\pi / 2) = 0$.Substituting the values: $A_R = \sqrt{(\sqrt{I})^2 + (2\sqrt{I})^2 + 2(\sqrt{I})(2\sqrt{I}) \cos(\pi / 2)}$.$A_R = \sqrt{I + 4I + 0} = \sqrt{5I}$.Since $A_1 = \sqrt{I} = A$,then $A_R = \sqrt{5}A$.

Two light waves of intensities $I$ and $4I$ superpose at a point with a phase difference of $\pi / 2$. Calculate the resultant amplitude at that point.

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