Question

If two waves, each of intensity $I$ having the same frequency but differing by a constant phase angle of $60°$ superpose at a certain point in space, then the intensity of the resultant wave is:

Answer 1

Step 1: Intensity

1. The energy of a wave is directly proportional to the square of its amplitude.
2. Therefore, the more the amplitude, the higher the energy of the wave.
3. The definition of intensity is the amount of energy passing through a unit area per unit of time.
4. Hence, an increase in amplitudes increases the intensity of the waves as well.

Step 2: Calculate the Intensity

Let $I_1$ be the intensity when both waves of intensity $I$ reach the same phase.

Let $I_2$ be the intensity when both waves reach the phase difference of $90°$.

$\therefore I_1=I_2=I$

Given the constant phase angle of $\varphi =60°$

From the formula of resultant intensity, we know that

$I_r=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

Upon substituting the values we get

$I_r=2I+2\sqrt{I^2}\cos 60°$

$=2I+2I\times \frac{1}{2}$

$=3I$

Hence, the intensity of the resultant wave is $3I$.