In the double-slit experiment of Fig. 35-10, the electric fields of the waves arriving at point P are given by

$$\begin{gathered} {\mathit{E}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu V}\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{sin}\left[\left(1.26\times {10}^{15}\right)t\right] \\ {\mathit{E}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu V}\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{sin}\left[\left(1.26\times {10}^{15}\right)t+39.6\mathrm{rad}\right] \end{gathered}$$

Where, time$\mathit{t}$is in seconds. (a) What is the amplitude of the resultant electric field at point P ? (b) What is the ratio of the intensity ${\mathit{I}}_{\mathit{P}}$at point P to the intensity ${\mathit{I}}_{\mathit{c}\mathit{e}\mathit{n}}$at the center of the interference pattern? (c) Describe where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. In a phasor diagram of the electric fields, (d) at what rate would the phasors rotate around the origin and (e) what is the angle between the phasors?

The amplitude of electric field for both waves is $2\mathrm{\mu V}/\mathrm{m}$,

The phase difference for both waves is $\varphi =39.6\text{rador}2269°$,

The angular frequency of both waves is $\omega =1.26\times {10}^{15}\mathrm{rad}/\mathrm{s}$,

The phase difference of fringe pattern varies with the path difference. For minimum phase difference path difference should be minimum and vice versa.

The resultant electric field at point P is given as:

$E=2E_0\cos \left(\frac{\varphi }{2}\right)$

Substitute all the values in equation.

$\begin{array}{rcl}E& =& 2(2\mathrm{\mu V}/\mathrm{m})\cos \left(\frac{2269°}{2}\right)\\ & =& 2.32\mathrm{\mu V}/\mathrm{m}\end{array}$

Therefore, the resultant electric field at point P is $\begin{array}{rcl}& & 2.32\mathrm{\mu V}/\mathrm{m}\end{array}$.

The ratio of intensity at point P to intensity at center of interference pattern is given as:

$\frac{I_P}{I_{cen}}=4{\cos }^2\left(\frac{\varphi }{2}\right)$

Substitute all the values in equation.

$$\begin{gathered} \frac{I_P}{I_{cen}}=4{\cos }^2\left(\frac{2269°}{2}\right) \\ \frac{I_P}{I_{cen}}=1.35 \end{gathered}$$

Therefore, the ratio of intensity at point P to intensity at center of interference pattern is $1.35$.

The phase difference for both waves in terms of wavelength of wave is given as:

${\varphi }_{\lambda }=\left(\frac{\varphi }{2\pi }\right)\lambda$

Substitute all the values in equation.

$\begin{array}{rcl}{\varphi }_{\lambda }& =& \left(\frac{39.6\text{rad}}{2\pi }\right)\lambda \\ & =& 6.3\lambda \end{array}$

The point P lies between sixth of maximum and seventh of minimum fringe.

Therefore, the point P is lying between sixth maxima and seventh minima of phasor diagram.

The rate of rotation of phasor will be equal to the angular frequency of both waves.

The rate of rotation of phasor is given as:

$\frac{d\varphi }{dt}=\omega$

Substitute all the values in the above equation.

$\frac{d\varphi }{dt}=1.26\times {10}^{15}\mathrm{rad}/\mathrm{s}$

Therefore, the rate of rotation of phasors is $1.26\times {10}^{15}\mathrm{rad}/\mathrm{s}$.

The angle between the phasors will be equal to the phase difference between both waves.

The angle between phasors is given as:

$\begin{array}{rcl}\theta & =& \varphi \\ & =& 39.6\text{rad}\\ & =& \left(39.6\text{rad}\right)\left(\frac{180°}{\pi }\right)\\ & =& 2269°\end{array}$

Therefore, the angle between phasors is$2269°$