Question

Two coherent sources emit light waves which superimpose at a point where these can be expressed as
$E_1=E_0\sin (\omega t+\frac{\pi }{4})$
$E_2=2E_0\sin (\omega t-\frac{\pi }{4})$
Here, $E_1$ and $E_2$ are the electric field strengths of the two waves at the given point. If I is the intensity of wave expressed by field strength $E_1$, find the resultant intensity (in $I$).

Answer 1

Given, $E_1=E_0\sin (\omega t+\frac{\pi }{4})$
$E_2=2E_0\sin (\omega t-\frac{\pi }{4})$
Here, $I\propto E^2\Rightarrow \frac{I_{result}}{I}=(\frac{E_{result}}{E}{)}^2......(1)$

at $t=0$
$E_{result}=\sqrt{E_1^2+E_2^2+2E_1{E}_{2}cos\theta }$
phase difference between $E_1$ and $E_2$ is $\frac{\pi }{2}.$
so, $\theta ={90}^o\to cos{90}^o=0$
hence, $E_{result}=\sqrt{E_1^2+E_2^2}$
where,$E_1=E_O,E_2=2E_0$

so, $E_{result}=\sqrt{E_0^2+(2E_0{)}^2}\Rightarrow =\sqrt{5}{E}_0$

using eq.(1)
$\frac{I_{result}}{I}=(\frac{E_{result}}{E}{)}^2\frac{I_{result}}{I}=\frac{5E_0}{E_0}{I}_{result}=5I$

So, $I_{result}=5$