Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0- -3- 2 -3 and - When they are superposed- the intensity of the resulting wave is nI0- The value of n is

I = I_0 + I_0 + 2 √(I_0)√(I_0)cos 60^∘ ⇒ I' = 3I_0 ⇒ n = 3 ...

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Standard IX

Physics

Frequency

Four harmonic...

Question

Four harmonic waves of equal frequencies and equal intensities $I_{0}$ have phase angles $0, \frac{π}{3}, \frac{2 π}{3}$ and $π$ . When they are superposed, the intensity of the resulting wave is $n I_{0}$ . The value of $n$ is

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Solution

$I = I_{0} + I_{0} + 2 \sqrt{I_{0}} \sqrt{I_{0}} cos 60^{\circ}$
$\Rightarrow I^{'} = 3 I_{0}$
$\Rightarrow n = 3$

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Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,π3,2π3 and π. When they are superposed, the intensity of the resulting wave is nI0. The value of n is

I=I0+I0+2√I0√I0cos60∘ ⇒I′=3I0 ⇒n=3

Four harmonic waves of equal frequencies and equal intensities $I_{0}$ have phase angles $0, \frac{π}{3}, \frac{2 π}{3}$ and $π$ . When they are superposed, the intensity of the resulting wave is $n I_{0}$ . The value of $n$ is

$I = I_{0} + I_{0} + 2 \sqrt{I_{0}} \sqrt{I_{0}} cos 60^{\circ}$
$\Rightarrow I^{'} = 3 I_{0}$
$\Rightarrow n = 3$