Chapter 37, Problem 74CP

To determine

To show that the equation $I=I_{\text{max}}\left(1+3{\cos }^2\varphi /2\right)$ is used for this situation.

Answer to Problem 74CP

It is shown that the equation $\underset{\_}{I=I_{\text{max}}\left(1+3{\cos }^2\varphi /2\right)}$ is used for this situation.

Explanation of Solution

Write the expression to find the magnitude of the light arriving at some point $P$.

    $\begin{array}{c}{E}_P=E_1+E_2\\ =3E_0\sin \omega t+E_0\sin \left(\omega t+\varphi \right)\\ =E_0\left(3\sin \omega t+\sin \left(\omega t+\varphi \right)\right)\end{array}$

Rewrite the above equation as ratio with maximum amplitude.

    $\begin{array}{c}\frac{E_p}{E_0}=3\sin \omega t+\sin \left(\omega t+\varphi \right)\\ =3\sin \omega t+\left(\sin \left(\omega t\right)\cos \left(\varphi \right)+\cos \left(\omega t\right)\sin \left(\varphi \right)\right)\\ =\sin \left(\omega t\right)\left(3+\cos \left(\varphi \right)\right)+\cos \left(\omega t\right)\sin \left(\varphi \right)\end{array}$

Write the square of the expression.

    $\begin{array}{c}{\left(\frac{E_p}{E_0}\right)}^2={\sin }^2\left(\omega t\right){\left(3+\cos \left(\varphi \right)\right)}^2+2\sin \left(\omega t\right)\cos \left(\omega t\right)\left(3+\cos \left(\varphi \right)\right)\sin \left(\varphi \right)+{\cos }^2\left(\omega t\right){\sin }^2\left(\varphi \right)\\ ={\sin }^2\left(\omega t\right){\left(3+\cos \left(\varphi \right)\right)}^2+\sin \left(2\omega t\right)\left(3+\cos \left(\varphi \right)\right)\sin \left(\varphi \right)+{\cos }^2\left(\omega t\right){\sin }^2\left(\varphi \right)\end{array}$

Take the time average of the above equation.

    $\begin{array}{c}{\left(\frac{E_p}{E_0}\right)}^2=\frac{1}{2}{\left(3+\cos \left(\varphi \right)\right)}^2+\frac{1}{2}{\sin }^2\left(\varphi \right)\\ =\frac{1}{2}\left(9+6\cos \left(\varphi \right)+{\cos }^2\left(\varphi \right)+{\sin }^2\left(\varphi \right)\right)\\ =\frac{1}{2}\left(10+6\cos \left(\varphi \right)\right)\\ =\frac{1}{2}\left(10+6\left(2{\cos }^2\left(\frac{\varphi }{2}\right)-1\right)\right)\\ =\frac{1}{2}\left(4+12{\cos }^2\left(\frac{\varphi }{2}\right)\right)\\ =2\left(1+3{\cos }^2\left(\frac{\varphi }{2}\right)\right)\end{array}$

Conclusion:

Write the expression for intensity.

    $\begin{array}{c}I=E_p^2\\ =2E_0^2\left(1+3{\cos }^2\left(\frac{\varphi }{2}\right)\right)\end{array}$

Write the expression for maximum intensity.

    $\begin{array}{c}{I}_{\text{max}}=2E_0^2\left(1+3{\cos }^2\left(0\right)\right)\\ =8E_0^2\end{array}$

Write the expression for the ratio of intensity to maximum intensity.

    $\begin{array}{c}\frac{I}{I_{\text{max}}}=\frac{2E_0^2\left(1+3{\cos }^2\left(\frac{\varphi }{2}\right)\right)}{8E_0^2}\\ =\frac{1}{4}\left(1+3{\cos }^2\left(\frac{\varphi }{2}\right)\right)\end{array}$

Rewrite the above equation to find the expression for intensity.

    $I=\frac{I_{\text{max}}}{4}\left(1+3{\cos }^2\left(\frac{\varphi }{2}\right)\right)$

Therefore, it is shown that the equation $\underset{\_}{I=I_{\text{max}}\left(1+3{\cos }^2\varphi /2\right)}$ is used for this situation.