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

Hint
In this question, we are asked to find the resultant two waves differing by a phase difference. We need to use the formula for the resultant intensity given by $I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $ where the phase difference between the 2 waves is given $60^\circ $.
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Here $I$ is the resultant intensity,
${I_1}$ and ${I_2}$ are the intensities of the 2 waves and the phase difference between them is $\phi $

Complete step by step answer
We can calculate the intensity of the resultant wave is calculated using the formula as,
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
In the given question the intensity of the 2 waves are said to be equal and are denoted by $I$,
$\Rightarrow {I_1} = {I_2} = I$
The frequency of the 2 waves are equal so we can write,
$\Rightarrow {f_1} = {f_2} = f$
In the question, we are given the phase difference between the 2 waves are,
$\Rightarrow \phi = 60^\circ $
Let us consider the intensity of the resultant wave to be represented by “${I_0}$”. Now substituting the given values in the formula we find the value of the intensity of the resultant wave as,
$\Rightarrow {I_0} = I + I + 2\sqrt {I \times I} \cos 60^\circ $
The value of $\cos 60^\circ $ is $\dfrac{1}{2}$
So by substituting the values we get,
$\Rightarrow {I_0} = I + I + 2\sqrt {I \times I} \times \dfrac{1}{2}$
Hence the third term after cancelling the 2 and removing the root has a value of $I$, So we get,
$\Rightarrow {I_0} = 2I + I$
That is,
$\Rightarrow {I_0} = 3I$
Therefore, the intensity of resultant wave of 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 is $3I$.

Additional Information
The resultant of the waves for a constructive interference will be maximum, as the value of the resultant intensity equals to ${I_0} = \sqrt {{{\left( {{I_1} + {I_2}} \right)}^2}} $.

Note
In this question we are given the phase difference between the 2 waves as $60^\circ $. The phase difference between 2 waves is the time difference between the same positions within the wave cycles. It is the difference in degrees when the 2 waves reach their maximum values.