Question

Two sources of intensity $I$ and $4I$ are used in an interference experiment, find intensity of a point where the wave from these two sources superimpose with a phase difference of $0,\dfrac{\pi }{2},\pi $.

Answer 1

Hint:This question can be solved by using the superposition principle which states that when two or more waves overlap in space, the resultant disturbance is equal to the algebraic sum of the individual disturbances. Here we will use the resultant intensity formula to find net intensity at a point.

Complete step by step answer:
According to the superposition principle we have the general formula for finding net intensity at a point due to two different intensities waves say $I$ and $I'$ and having a phase difference of $\phi $ . Net intensity due to both waves is given by ${I_{net}} = I + I' + 2\sqrt {I \times I'} \cos \phi $ .
Now we will find net intensity for given two waves of intensities $I$ and $4I$ , having a phase difference of $0$. So, $\cos {0^ \circ } = 1$ and net intensity is given by,
${I_{net}} = I + 4I + 2\sqrt {I \times 4I} \\
\Rightarrow {I_{net}} = 5I + 4I \\
\Rightarrow {I_{net}} = 9I \\ $
Now, we will find for phase difference of $\dfrac{\pi }{2}$
So, $\cos \dfrac{\pi }{2} = 0$ , net intensity is given by
${I_{net}} = I + 4I + 2\sqrt {I \times 4I} \times 0 \\
\Rightarrow {I_{net}} = 5I \\ $
Now, we will find for phase difference of $\pi $
So, $\cos \pi = - 1$ , net intensity is given by
${I_{net}} = I + 4I - 2\sqrt {I \times 4I} \\
\Rightarrow {I_{net}} = 5I - 4I \\
\therefore {I_{net}} = I \\ $
Hence, the net intensity for phase difference of $0$ is $9I$ phase difference for $\dfrac{\pi }{2}$ is $5I$ and for phase difference of $\pi $ is $I$.

Note: Remember, Interference is the phenomenon in which two waves superpose to each other to form the resultant wave of either lower or higher and even the same amplitude. Whenever the resultant amplitude gets increased from initial amplitude it’s called constructive Interference and when it gets decreased it’s called destructive interference.