Question

Two coherent sources emit light waves which superimpose at a point and are expressed as $\begin{array}{c}{E}_1=E_0\sin (\omega t+\pi /4)\\ E_2=2E_0\sin (\omega t-\pi /4)\end{array}$
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 $E_1$ . The resultant intensity istimes $I$.

Answer 1

Intensity of the wave expressed by field strength $E_1$ is
$\begin{array}{c}I\propto E_0^2\\ \text{As [intensity}\propto {\left(\mathrm{amplitude}\right)}^2\text{]}\end{array}$
Intensity of the wave expressed by field strength $E_2$ is
$\begin{array}{c}{I}^{\prime }\propto {\left(2E_0\right)}^2\\ \frac{I^{\prime }}{I}=4\text{or}{I}^{\prime }=4I\end{array}$
where $I$ is the intensity of wave expressed by field strength $E_1$.
The phase difference between the two waves is given by
$\varphi =(\omega t+\pi /4)-(\omega t-\pi /4)=\frac{\pi }{2}$
$\therefore$ Resultant intensity is given by
$I_R=I+I\text{'}+2\sqrt{I{I}^{\prime }}\cos \varphi \Rightarrow \left(I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi \right]\Rightarrow I_R=I+4I+2\sqrt{I\left(4I\right)}\cos \pi /2\Rightarrow I_R=5I$
Thus, resultant intensity is five times the intensity of wave expressed by $E_1$