Question

Three waves from three coherent sources meet at the same point. The Amplitude of each wave is $A_0$. Intensity corresponding to $A_0$ is $I_0$. At a point, the phase difference between the first wave and the second wave is ${60}^{\circ }$and the path difference between the first wave and the third wave is $\frac{\lambda }{3}$. The first wave lags behind in phase angle from the second and the third wave. The resultant intensity at this point is

• A. $I_0$
• B. $2I_0$
• C. $3I_0$
• D. $4I_0$

Answer 1

The correct option is D $4I_0$

Let us assume first wave as reference.

Now,
Phase difference between first wave and second wave, ${\varphi }_{21}={60}^{\circ }$

Path difference between the first wave and the third wave, $\Delta {\text{x}}_{31}=\frac{\lambda }{3}$.

$\Rightarrow \Delta {\varphi }_{31}=\frac{2\pi }{\lambda }\times \Delta {\text{x}}_{31}=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}$

$\therefore \Delta {\varphi }_{31}=\frac{2\pi }{3}={120}^{\circ }$

So, a vector diagram for amplitudes is shown in figure.


Net magnitude of two equal vectors (here, $A_1=A_3=A_0$) passes through angle bisector (here in direction of $A_2$).

Also, as the angle between $\overrightarrow{A_1}$ and $\overrightarrow{A_3}\text{is}{120}^{\circ }$, the net magnitude of these two vectors will be $A_0$.

So, the net magnitude, $A_{net}=A_0+A_0=2A_0$.

So, the resultant intensity, $I=4I_0(\text{as}I\propto A^2)$

Hence, option $\left(D\right)$ is correct.

Why this question? Key note: For vector addition of more than two vectors, add two vectors at a time and then add its resultant to the third vector.