Q17E - University Physics with Modern Physics

Question

In a two-slit interference pattern, the intensity at the peak of the central maximum is ${\mathit{I}}_{\mathbf{0}}$. (a) At a point in the pattern where the phase difference between the waves from the two slits is $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$, what is the intensity? (b) What is the path difference for $\mathbf{480}\mathbf{-}\mathit{n}\mathit{m}$ light from the two slits at a point where the phase difference is $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$?

Step-by-Step Solution

Answer

1. The intensity is $\mathbf{0}\mathbf{.}\mathbf{75}{\mathit{I}}_{\mathbf{0}}$.
2. The path difference from the two slits is $\mathbf{80}\mathit{n}\mathit{m}$.

Step 1: Formulas used to solve the question

Intensity is given by

                                       $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathbf{ }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\varphi }{\mathbf{2}}\mathbf{\right)}$

Phase difference is given by

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

Step 2: Calculate the intensity

Given:            $\mathit{\varphi }\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{3}}\mathit{r}\mathit{a}\mathit{d}$

            $\mathit{\lambda }\mathbf{=}\mathbf{480}\mathit{n}\mathit{m}\mathbf{=}\mathbf{480}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathit{m}$

Now, the intensity is given by

                                              $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathbf{ }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\varphi }{\mathbf{2}}\mathbf{\right)}$

Plug the given,

                                           $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathbf{ }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{60}}{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}{\mathit{I}}_{\mathbf{0}}$

Step 3: Calculate the path difference

From equation (2), solve for $\mathbf{∆}\mathit{r}$

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

Plug the given,

                              $\mathbf{∆}\mathit{r}\mathbf{=}\frac{\mathbf{480}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{*}{\displaystyle \frac{\mathbf{\pi }}{\mathbf{3}}}}{\mathbf{2}\mathbf{\pi }}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathit{m}\mathbf{=}\mathbf{80}\mathit{n}\mathit{m}$

Thus, the intensity is $\mathbf{0}\mathbf{.}\mathbf{75}{\mathit{I}}_{\mathbf{0}}$. The path difference from the two slits is $\mathbf{80}\mathit{n}\mathit{m}$.