Question

Two monochromatic sinusoidal waves, each of intensity $I$, have a constant phase difference of $\varphi$. These waves when superimposed, form a resultant sinusoidal wave. Then, the intensity of the resultant wave is

• A. $4I$
• B. $4I\cos \varphi$
• C. $4I{\cos }^2\varphi$
• D. $4I{\cos }^2\left(\frac{\varphi }{2}\right)$

Answer 1

The correct option is D $4I{\cos }^2\left(\frac{\varphi }{2}\right)$

From the data given in the question,
$I_1=I_2=I$
Phase difference $\left(\delta \right)=\varphi$


We know that,

$A_{net}^2={\left(A_1\right)}^2+{\left(A_2\right)}^2+2\left(A_1\right)\left(A_2\right)\cos \delta ......\left(1\right)$

Also, $I\propto A^2$

Therefore, $\left(1\right)$ can be rewritten as,

$I_{net}=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \delta$
From the given data,
$I_{net}=I+I+2\sqrt{I}\sqrt{I}\cos \varphi$
$\Rightarrow I_{net}=2I+2I\cos \varphi =2I\times 2{\cos }^2\left(\frac{\varphi }{2}\right)$
$[\text{Since},1+\cos \varphi =2{\cos }^2\left(\frac{\varphi }{2}\right)]$
$\Rightarrow I_{net}=4I{\cos }^2\left(\frac{\varphi }{2}\right)$

Thus, option (d) is the correct answer.