Q18E - University Physics with Modern Physics

Question

Coherent sources A and B emit electromagnetic waves with wavelength 2.00cm . Point P is 4.86m from A and 5.24m from B. What is the phase difference at P between these two waves?

Step-by-Step Solution

Answer

The phase difference between the two waves at point P is $38\pi ad$.

Step 1: Formula for phase difference

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{∆}\mathbf{r}\mathbf{*}\mathbf{+}\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

Step 2: Calculate the phase difference

Given:            $$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{\lambda }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{m} \\ {\mathit{r}}_{\mathbf{P}\mathbf{\to }\mathbf{A}}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{86}\mathit{m} \\ {\mathit{r}}_{\mathbf{P}\mathbf{\to }\mathbf{B}}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{86}\mathit{m} \end{gathered}$$

         

Now, the phase difference is given by

                                     $$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{\varphi }\mathit{=}\frac{\mathit{∆}\mathit{r}\mathit{*}\mathit{2}\mathit{\pi }}{\mathit{\lambda }} \\ \mathit{\Rightarrow }\mathit{\varphi }\mathit{=}\frac{\mathit{∆}\left(r_2-r_1\right)\mathit{\pi }}{\mathit{\lambda }} \end{gathered}$$

Plug the given,

                                   $\mathit{\varphi }\mathbf{=}\frac{\left(5.24-4.86\right)}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}\mathbf{=}\mathbf{38}\mathit{\pi }\mathit{r}\mathit{a}\mathit{d}$

Thus, the phase difference between the two waves at point P is $\mathit{38}\mathit{\pi }\mathit{a}\mathit{d}$ .