Question

Two waves of sound having intensities $I$ and $4I$ interfere to produce interference pattern. The phase difference between the waves is $\pi /2$ at point A and $\pi$ at point B. Then the difference between the resultant intensities at A and B is

• A. 2I
• B. 4I
• C. 5I
• D. 7I

Answer 1

The correct option is B 4I

$A_1/A_2=\sqrt{I_1/I_2}=1/2$,
say,

$Y_1=A\sin (\omega \ast t)$, $Y_2=2A\sin (\omega \ast t+\pi /2)$

by the law of superposition of waves,

At point A,

$Y=Y_1+Y_2=A\sin \omega \ast t+2A\cos \omega \ast t=\sqrt{5}A(\sin (\omega \ast t+\varphi ))$
here,

$\varphi ={\cos }^{-1}1/\sqrt{5}={\sin }^{-1}2/\sqrt{5}$

and $I_A=Y_A^2=5A^2{\sin }^2(\omega \ast t+\varphi )=5I$

At point B,

$Y=Y_1+Y_2=A\sin \omega \ast t+2A\sin (\omega \ast t+\pi )=A\sin \omega \ast t-2A\sin \omega \ast t=-A\sin \omega \ast t$

here, $I_B=A^2{\sin }^2\omega \ast t=I$
Option "B" is correct.