Four harmonic waves of equal frequencies and equal intensities $I_{0}$ have phase angles $0, π / 3,$ $2 π / 3$ 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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answer is 3.

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Detailed Solution

$\begin{array}{l} W a v e s w i t h p h a s e a n g l e s 0 a n d π, w i l l c a n c e l e a c h o t h e r . \\ R e m a i n i n g t w o w a v e s h a s p h a s e d i f f e r e n c e (△ ϕ) = \frac{2 π}{3} - \frac{π}{3} = \frac{π}{3} = 60^{0} \\ R e s u l \tan t I n t e n s i t y = I_{Res} = 4 I_{0} \cos^{2} (\frac{△ ϕ}{2}) \\ \Rightarrow I_{Res} = 4 I_{0} \cos^{2} (30^{\circ}) \\ \Rightarrow I_{Res} = 4 I_{0} \times \frac{3}{4} = 3 I_{0} \\ S o, n = 3 \end{array}$